Integrals: Approximation

Fall 2021.2 #real_analysis/qual/completed

  • Let \(F \subset \mathbb{R}\) be closed, and define \begin{align*} \delta_{F}(y):=\inf _{x \in F}|x-y| . \end{align*} For \(y \notin F\), show that \begin{align*} \int_{F}|x-y|^{-2} d x \leq \frac{2}{\delta_F(y)}, \end{align*}

  • Let \(F \subset \mathbb{R}\) be a closed set whose complement has finite measure, i.e. \(m({\mathbf{R}}\setminus F)< \infty\). Define the function \begin{align*} I(x):=\int_{\mathbb{R}} \frac{\delta_{F}(y)}{|x-y|^{2}} d y \end{align*} Prove that \(I(x)=\infty\) if \(x \not\in F\), however \(I(x)<\infty\) for almost every \(x \in F\).

    Hint: investigate \(\int_{F} I(x) d x\).

Let \(y\in F^c\) which is open, then one can find an epsilon ball about \(y\) avoiding \(F\). We can take \({\varepsilon}\coloneqq\delta_F(y)\) to define \(A \coloneqq B_{{\varepsilon}}(y)\), and we still have \(A \subseteq F^c\) and \(F \subseteq A^c\). Note that \({\left\lvert {x-y} \right\rvert}^2 = (x-y)^2\) since this is always positive, then \begin{align*} \int_F {\left\lvert {x-y} \right\rvert}^{-2} \,dx &\leq \int_{A^c} {\left\lvert {x-y} \right\rvert}^{-2} \,dx\\ &= \int_{-\infty}^{-{\varepsilon}} \qty{x-y}^{-2} \,dx+ \int_{{\varepsilon}}^{\infty} \qty{x-y}^{-2}\,dx\\ &= \int_{-\infty}^{-{\varepsilon}} u^{-2} \,dx+ \int_{{\varepsilon}}^{\infty} u^{-2} \,dx\\ &= -u^{-1}\Big|_{u=-{\varepsilon}}^{u=-\infty}- u^{-1}\Big|_{u=\infty}^{u={\varepsilon}} \\ &= {2\over {\varepsilon}} \\ &\coloneqq{2\over \delta_F(y)} .\end{align*}

Estimate: \begin{align*} \int_F I(x) \,dx &\coloneqq\int_F \int_{\mathbf{R}}{\delta_F(y) \over (x-y)^2 } \,dy\,dx\\ &= \int_{\mathbf{R}}\delta_F(y) \int_F {1\over (x-y)^2} \,dx\,dy\\ &= \int_F \delta_F(y) \int_F {1\over (x-y)^2} \,dx\,dy+ \int_{F^c} \delta_F(y) \int_F {1\over (x-y)^2} \,dx\,dy\\ &= 0 + \int_{F^c} \delta_F(y) \int_F {1\over (x-y)^2} \,dx\,dy\\ &\leq \int_{F^c} 2 \,dy\\ &= 2\mu(F^c) \\ &<\infty ,\end{align*} where we’ve used that \(y\in F\implies \delta_F(y) = 0\) and applied the bound from the first part. We’ve also implicitly used Fubini-Tonelli to change the order of integration, justified by positivity of the integrand and the finite iterated integral. This forces \(I(x) < \infty\) for almost every \(x\in F\), since if \(I(x)\) is unbounded on any positive measure set then this integral would diverge.

If \(x\not\in F\), pick an \({\varepsilon}{\hbox{-}}\)ball \(A\) about \(x\) avoiding \(F\) so that \({\left\lvert {x-y} \right\rvert}> {\varepsilon}\) for any \(y\in A^c\) and thus \((x-y)^{-2} \leq {\varepsilon}^{-2}\). Note that \(\delta_F(y)\geq {\varepsilon}\), so \begin{align*} I(x) &= \int_{\mathbf{R}}\delta_F(y) (x-y)^{-2} \,dy\\ &\geq \int_{A^c} \delta_F(y) (x-y)^{-2} \,dy\\ &\geq \int_{A^c} \delta_F(y) {\varepsilon}^{-2} \,dy\\ &\geq \int_{A^c} {\varepsilon}^{-1} \,dy\\ &= \mu(A^c){\varepsilon}^{-1} ,\end{align*} which can be made arbitrarily large by taking \({\varepsilon}\to 0\).

#todo: Not great, \(A^c\) depends on \({\varepsilon}\) so this ratio has a competing numerator…

Spring 2018.3 #real_analysis/qual/completed

Let \(f\) be a non-negative measurable function on \([0, 1]\).

Show that \begin{align*} \lim _{p \rightarrow \infty}\left(\int_{[0,1]} f(x)^{p} d x\right)^{\frac{1}{p}}=\|f\|_{\infty}. \end{align*}

  • \({\left\lVert {f} \right\rVert}_\infty \coloneqq\inf_t {\left\{{ t{~\mathrel{\Big\vert}~}m\qty{\left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}f(x) > t}\right\}} = 0 }\right\} }\), i.e. this is the lowest upper bound that holds almost everywhere.

  • \({\left\lVert {f} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_\infty\):
    • Note \({\left\lvert {f(x)} \right\rvert} \leq {\left\lVert {f} \right\rVert}_\infty\) almost everywhere and taking \(p\)th powers preserves this inequality.

    • Thus \begin{align*} {\left\lvert {f(x)} \right\rvert} &\leq {\left\lVert {f} \right\rVert}_\infty \quad\text{a.e. by definition} \\ \implies {\left\lvert {f(x)} \right\rvert}^p &\leq {\left\lVert {f} \right\rVert}_\infty^p \quad\text{for } p\geq 0 \\ \implies {\left\lVert {f} \right\rVert}_p^p &= \int_X {\left\lvert {f(x)} \right\rvert}^p ~dx \\ &\leq \int_X {\left\lVert {f} \right\rVert}_\infty^p ~dx \\ &= {\left\lVert {f} \right\rVert}_\infty^p \int_X 1\,dx \\ &= {\left\lVert {f} \right\rVert}_\infty^p \cdot m(X) \quad\text{since the norm doesn't depend on }x \\ &= {\left\lVert {f} \right\rVert}_\infty^p \qquad \text{since } m(X) = 1 .\end{align*}

      • Thus \({\left\lVert {f} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_\infty\) for all \(p\) and taking \(\lim_{p\to\infty}\) preserves this inequality.
  • \({\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty\):
    • Fix \(\varepsilon > 0\).

    • Define \begin{align*} S_\varepsilon \coloneqq\left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} \geq {\left\lVert {f} \right\rVert}_\infty - \varepsilon}\right\} .\end{align*}

      • Note that \(m(S_{\varepsilon}) > 0\); otherwise if \(m(S_{\varepsilon}) = 0\), then \(t\coloneqq{\left\lVert {f} \right\rVert}_\infty - {\varepsilon}< {\left\lVert {f} \right\rVert}_{\varepsilon}\). But this produces a smaller upper bound almost everywhere than \({\left\lVert {f} \right\rVert}_{\varepsilon}\), contradicting the definition of \({\left\lVert {f} \right\rVert}_{\varepsilon}\) as an infimum over such bounds.
    • Then \begin{align*} {\left\lVert {f} \right\rVert}_p^p &= \int_X {\left\lvert {f(x)} \right\rvert}^p ~dx \\ &\geq \int_{S_\varepsilon} {\left\lvert {f(x)} \right\rvert}^p ~dx \quad\text{since } S_{\varepsilon}\subseteq X \\ &\geq \int_{S_\varepsilon} {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert}^p ~dx \quad\text{since on } S_{\varepsilon}, {\left\lvert {f} \right\rvert} \geq {\left\lVert {f} \right\rVert}_\infty - {\varepsilon}\\ &= {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert}^p \cdot m(S_\varepsilon) \quad\text{since the integrand is independent of }x \\ & \geq 0 \quad\text{since } m(S_{\varepsilon}) > 0 \end{align*}

    • Taking \(p\)th roots for \(p\geq 1\) preserves the inequality, so \begin{align*} \implies {\left\lVert {f} \right\rVert}_p &\geq {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert} \cdot m(S_\varepsilon)^{\frac 1 p} \overset{p\to\infty}\longrightarrow{\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert} \overset{\varepsilon \to 0}\longrightarrow{\left\lVert {f} \right\rVert}_\infty \end{align*} where we’ve used the fact that above arguments work

    • Thus \({\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty\).

Spring 2018.4 #real_analysis/qual/completed

Let \(f\in L^2([0, 1])\) and suppose \begin{align*} \int _{[0,1]} f(x) x^{n} d x=0 \text { for all integers } n \geq 0. \end{align*} Show that \(f = 0\) almost everywhere.

  • Weierstrass Approximation: A continuous function on a compact set can be uniformly approximated by polynomials.
  • \(C^1([0, 1])\) is dense in \(L^2([0, 1])\)
  • Polynomials are dense in \(L^p(X, \mathcal{M}, \mu)\) for any \(X\subseteq {\mathbf{R}}^n\) compact and \(\mu\) a finite measure, for all \(1\leq p < \infty\).
    • Use Weierstrass Approximation, then uniform convergence implies \(L^p(\mu)\) convergence by DCT.

  • Fix \(k \in {\mathbf{Z}}\).

  • Since \(e^{2\pi i k x}\) is continuous on the compact interval \([0, 1]\), it is uniformly continuous.

  • Thus there is a sequence of polynomials \(P_\ell\) such that \begin{align*} P_{\ell, k} \overset{\ell\to\infty}\longrightarrow e^{2\pi i k x} \text{ uniformly on } [0,1] .\end{align*}

  • Note applying linearity to the assumption \(\int f(x) \, x^n\), we have \begin{align*} \int f(x) x^n \,dx = 0 ~\forall n \implies \int f(x) p(x) \,dx = 0 \end{align*} for any polynomial \(p(x)\), and in particular for \(P_{\ell, k}(x)\) for every \(\ell\) and every \(k\).

  • But then
    \begin{align*} {\left\langle {f},~{e_k} \right\rangle} &= \int_0^1 f(x) e^{-2\pi i k x} ~dx \\ &= \int_0^1 f(x) \lim_{\ell \to \infty} P_\ell(x) \\ &= \lim_{\ell \to \infty} \int_0^1 f(x) P_\ell(x) \quad\quad \text{by uniform convergence on a compact interval} \\ &= \lim_{\ell \to \infty} 0 \quad\text{by assumption}\\ &= 0 \quad \forall k\in {\mathbf{Z}} ,\end{align*} so \(f\) is orthogonal to every \(e_k\).

  • Thus \(f\in S^\perp \coloneqq\mathop{\mathrm{span}}_{\mathbf{C}}\left\{{e_k}\right\}_{k\in {\mathbf{Z}}}^\perp \subseteq L^2([0, 1])\), but since this is a basis, \(S\) is dense and thus \(S^\perp = \left\{{0}\right\}\) in \(L^2([0, 1])\).

  • Thus \(f\equiv 0\) in \(L^2([0, 1])\), which implies that \(f\) is zero almost everywhere.

  • By density of polynomials, for \(f\in L^2([0, 1])\) choose \(p_{\varepsilon}(x)\) such that \({\left\lVert {f - p_{\varepsilon}} \right\rVert} < {\varepsilon}\) by Weierstrass approximation.

  • Then on one hand, \begin{align*} {\left\lVert {f(f-p_{\varepsilon})} \right\rVert}_1 &= {\left\lVert {f^2} \right\rVert}_1 - {\left\lVert {f\cdot p_{\varepsilon}} \right\rVert}_1 \\ &= {\left\lVert {f^2} \right\rVert}_1 - 0 \quad\text{by assumption} \\ &= {\left\lVert {f} \right\rVert}_2^2 .\end{align*}

    • Where we’ve used that \({\left\lVert {f^2} \right\rVert}_1 = \int {\left\lvert {f^2} \right\rvert} = \int {\left\lvert {f} \right\rvert}^2 = {\left\lVert {f} \right\rVert}_2^2\).
  • On the other hand \begin{align*} {\left\lVert {f(f-p_{\varepsilon})} \right\rVert} &\leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {f-p_{\varepsilon}} \right\rVert}_\infty \quad\text{by Holder} \\ &\leq {\varepsilon}{\left\lVert {f} \right\rVert}_1 \\ &\leq {\varepsilon}{\left\lVert {f} \right\rVert}_2 \sqrt{m(X)} \\ &= {\varepsilon}{\left\lVert {f} \right\rVert}_2 \quad\text{since } m(X)= 1 .\end{align*}

    • Where we’ve used that \({\left\lVert {fg} \right\rVert}_1 = \int {\left\lvert {fg} \right\rvert} = \int {\left\lvert {f} \right\rvert}{\left\lvert {g} \right\rvert} \leq \int {\left\lVert {f} \right\rVert}_\infty {\left\lvert {g} \right\rvert} = {\left\lVert {f} \right\rVert}_\infty {\left\lVert {g} \right\rVert}_1\).
  • Combining these, \begin{align*} {\left\lVert {f} \right\rVert}_2^2 \leq {\left\lVert {f} \right\rVert}_2 {\varepsilon}\implies {\left\lVert {f} \right\rVert}_2 < {\varepsilon}\to 0, .\end{align*} so \({\left\lVert {f} \right\rVert}_2 = 0\), which implies \(f=0\) almost everywhere.

Spring 2015.2 #real_analysis/qual/work

Let \(f: {\mathbf{R}}\to {\mathbf{C}}\) be continuous with period 1. Prove that \begin{align*} \lim _{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^{N} f(n \alpha)=\int_{0}^{1} f(t) d t \quad \forall \alpha \in {\mathbf{R}}\setminus{\mathbf{Q}}. \end{align*}

Hint: show this first for the functions \(f(t) = e^{2\pi i k t}\) for \(k\in {\mathbf{Z}}\).

Fall 2014.4 #real_analysis/qual/completed

Let \(g\in L^\infty([0, 1])\) Prove that \begin{align*} \int _{[0,1]} f(x) g(x)\, dx = 0 \quad\text{for all continuous } f:[0, 1] \to {\mathbf{R}} \implies g(x) = 0 \text{ almost everywhere. } \end{align*}

  • Polar decomposition: \(f = \operatorname{sign}(f) \cdot {\left\lvert {f} \right\rvert}\).
  • \(L^\infty[0, 1] \subseteq L^1[0, 1]\).

Use that \(L^\infty[0, 1] \subseteq L^1[0, 1]\), so fixing \(g\), choose a sequence of compactly supported continuous functions \(f_k\) converging to \(\operatorname{sign}(g)\) in \(L^1\). We can arrange so that \({\left\lvert {g_k} \right\rvert} \leq 1\). Then \begin{align*} \int {\left\lvert {g} \right\rvert} &= \int\operatorname{sign}(g)\cdot g \\ &= \int \lim_k g_k\cdot g \\ &\overset{\text{DCT}}{=} \lim_k \int g_k\cdot g \\ &=\lim_k 0 \\ &= 0 ,\end{align*} where the DCT applies since defining \(h_k \coloneqq g_k\cdot g\) we have \({\left\lvert {h_k} \right\rvert} \leq g\in L^1[0, 1]\), and each integral is zero since \(g_k\) is continuous (and we use the hypothesis).


Spring 2021.4 #real_analysis/qual/completed

Let \(f, g\) be Lebesgue integrable on \({\mathbf{R}}\) and let \(g_n(x) \coloneqq g(x- n)\). Prove that \begin{align*} \lim_{n\to \infty } {\left\lVert {f + g_n} \right\rVert}_1 = {\left\lVert {f} \right\rVert}_1 + {\left\lVert {g} \right\rVert}_1 .\end{align*}

  • For \(f\in L^1(X)\), \({\left\lVert {f} \right\rVert}_1 \coloneqq\int_X {\left\lvert {f(x)} \right\rvert} \,dx< \infty\).

  • Small tails in \(L_1\): if \(f\in L^1({\mathbf{R}}^n)\), then for every \({\varepsilon}>0\) exists some radius \(R\) such that \begin{align*} {\left\lVert {f} \right\rVert}_{L^1(B_R^c)} < {\varepsilon} .\end{align*}

  • Shift \(g\) to the right far enough so that the two densities are mostly disjoint:

Shifting density

  • Any integral \(\int_a^b f\) can be written as \({\left\lVert {f} \right\rVert}_1 - O(\text{err})\).

  • Bounding technique: \begin{align*} a-{\varepsilon}\leq b \leq a+{\varepsilon}\implies b=a .\end{align*}

  • Fix \({\varepsilon}\).

  • Using small tails for \(f, g \in L^1\), choose \(R_1, R_2 \gg 0\) so that \begin{align*} \int_{B_{R_1}(0)^c} {\left\lvert {f} \right\rvert} &< {\varepsilon}\\ \int_{B_{R_2}(0)^c} {\left\lvert {g} \right\rvert} &< {\varepsilon} .\end{align*}

    • Note that this implies \begin{align*} \int_{-R_1}^{R_1} {\left\lvert {f} \right\rvert} &= {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon}\\ \int_{-R_2}^{R_2} {\left\lvert {g_N} \right\rvert} &= {\left\lVert {g_N} \right\rVert} - 2{\varepsilon} .\end{align*}

    • Also note that by translation invariance of the Lebesgue integral, \({\left\lVert {g} \right\rVert}_1 = {\left\lVert {g_N} \right\rVert}_1\).

  • Now use \(N\) to make the densities almost disjoint: choose \(N\gg 1\) so that \(N-R_2 > R_1\):

Shifting density

  • Consider the change of variables \(x\mapsto x-N\): \begin{align*} \int_{-R_2}^{R_2} {\left\lvert {g(x)} \right\rvert}\,dx = \int_{N-R_2} ^{N+R_2} {\left\lvert {g(x-N)} \right\rvert} \,dx \coloneqq\int_{N-R_2} ^{N+R_2} {\left\lvert {g_N(x)} \right\rvert} \,dx .\end{align*}
    • Use this to conclude that \begin{align*} \int_{N-R_2}^{N+R_2} {\left\lvert {g_N} \right\rvert} = {\left\lVert {g_N} \right\rVert} - 2{\varepsilon} .\end{align*}
  • Now split the integral in the problem statement at \(R_1\):

\begin{align*} {\left\lVert {f + g_N} \right\rVert}_1 = \int_{\mathbf{R}}{\left\lvert {f+g_N} \right\rvert} = \int_{-\infty}^{R_1} {\left\lvert {f+ g_N} \right\rvert} + \int_{R_1}^{\infty} {\left\lvert {f+ g_N} \right\rvert} \coloneqq I_1 + I_2 .\end{align*}

  • Idea: from the picture,

    • On \(I_1\), \(f\) is big and \(g_N\) is small
    • On \(I_2\), \(f\) is small and \(g_N\) is big
  • Casework: estimate \(I_1, I_2\) separately, bounding from above and below.

  • \(I_1\) upper bound: \begin{align*} I_1 &\coloneqq\int_{-\infty}^{R_1} {\left\lvert {f + g_N} \right\rvert} \\ &\leq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} + {\left\lvert {g_N} \right\rvert} \\ &= \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} + \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\leq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} + \int_{-\infty}^{\color{green} N - R_2} {\left\lvert {g_N} \right\rvert} && R_1 < N-R_2 \\ &= {\left\lVert {f} \right\rVert}_1 - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + \int_{-\infty}^{N - R_2} {\left\lvert {g_N} \right\rvert} \\ &\leq {\left\lVert {f} \right\rVert}_1 - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + {\varepsilon}\\ &\leq {\left\lVert {f} \right\rVert}_1 + {\varepsilon} .\end{align*}

    • In the last step we’ve used that we’re subtracting off a positive number, so forgetting it only makes things larger.

    • We’ve also used monotonicity of the Lebesgue integral: if \(A\leq B\), then \((c, A) \subseteq (c, B)\) and \(\int_{c}^A {\left\lvert {f} \right\rvert} \leq \int_c^B {\left\lvert {f} \right\rvert}\) since \({\left\lvert {f} \right\rvert}\) is positive.

  • \(I_1\) lower bound: \begin{align*} I_1 &\coloneqq\int_{-\infty}^{R_1} {\left\lvert {f + g_N} \right\rvert} \\ &\geq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} - {\left\lvert {g_N} \right\rvert} \\ &= \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\geq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} - \int_{-\infty}^{\color{green} N-R_2} {\left\lvert {g_N} \right\rvert} && R_1 < N-R_2 \\ &= {\left\lVert {f} \right\rVert}_1 - \int_{R_1}^{ \infty } {\left\lvert {f} \right\rvert} - \int_{- \infty }^{N-R_2} {\left\lvert {g_N} \right\rvert} \\ &\geq {\left\lVert {f} \right\rVert}_1 - {\varepsilon}- {\varepsilon}\\ &= {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon} .\end{align*}

    • Now we’ve used that the integral with \(g_N\) comes in with a negative sign, so extending the range of integration only makes things smaller. We’ve also used the \({\varepsilon}\) bound on both \(f\) and \(g_N\) here, and both are tail estimates.
  • Taken together we conclude \begin{align*} {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon} \leq I_1 \leq {\left\lVert {f} \right\rVert}_1 && {\varepsilon}\to 0 \implies I_1 = {\left\lVert {f} \right\rVert}_1 .\end{align*}

  • \(I_2\) lower bound: \begin{align*} I_2 &\coloneqq\int_{R_1}^{\infty} {\left\lvert {f + g_N} \right\rvert} \\ &\leq \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + \int_{R_1}^{\infty} {g_N} \\ &\leq \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + {\left\lVert {g_N} \right\rVert}_1 - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\leq {\varepsilon}+ {\left\lVert {g_N} \right\rVert}_1 - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\leq {\varepsilon}+ {\left\lVert {g_N} \right\rVert}_1 \\ &= {\varepsilon}+ {\left\lVert {g} \right\rVert}_1 .\end{align*}

    • Here we’ve again thrown away negative terms, only increasing the bound, and used the tail estimate on \(f\).
  • \(I_2\) upper bound:

\begin{align*} I_2 &\coloneqq\int_{R_1}^{\infty} {\left\lvert {f + g_N} \right\rvert} \\ &= \int_{R_1}^{\infty} {\left\lvert {g_N + f} \right\rvert} \\ &\geq \int_{R_1}^{\infty} {\left\lvert {g_N} \right\rvert} - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} \\ &= {\left\lVert {g_N} \right\rVert} - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} \\ &\geq {\left\lVert {g_N} \right\rVert} - 2{\varepsilon} .\end{align*}

  • Here we’ve swapped the order under the absolute value, and used the tail estimates on both \(g\) and \(f\).

  • Taken together: \begin{align*} {\left\lVert {g} \right\rVert}_1 - {\varepsilon}\leq I_2 \leq {\left\lVert {g} \right\rVert}_1 + 2{\varepsilon} .\end{align*}

  • Note that we have two inequalities: \begin{align*} {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon}&\leq \int_{-\infty}^{R_1} {\left\lvert {f -g_N} \right\rvert} \leq {\left\lVert {f} \right\rVert}_1 + {\varepsilon}\\ {\left\lVert {g} \right\rVert}_1 - 2{\varepsilon}&\leq \int^{\infty}_{R_1} {\left\lvert {f -g_N} \right\rvert} \leq {\left\lVert {g} \right\rVert}_1 + {\varepsilon} .\end{align*}

  • Add these to obtain \begin{align*} {\left\lVert {f} \right\rVert}_1 + {\left\lVert {g} \right\rVert}_1 - 4{\varepsilon}\leq I_1 + I_2 \coloneqq{\left\lVert {f - g_N} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert} + {\left\lVert {g} \right\rVert}_1 + 2{\varepsilon} .\end{align*}

  • Check that as \(N\to \infty\) as \({\varepsilon}\to 0\) to yield the result.

Fall 2020.4 #real_analysis/qual/completed

Prove that if \(xf(x) \in L^1({\mathbf{R}})\), then \begin{align*} F(y) \coloneqq\int f(x) \cos(yx)\, dx \end{align*} defines a \(C^1\) function.

  • Fix \(y_0\), we’ll show \(F'\) exists and is continuous at \(y_0\).

  • Fix a sequence \(y_n\searrow y_0\) and define \begin{align*} h_n(x) \coloneqq { h(x, y_n) - h(x, y_0) \over y_n - y_0} && h(x, y) \coloneqq f(x) \cos(yx) .\end{align*}

  • We can then write \begin{align*} {\frac{\partial h}{\partial y}\,}(x, y_0) = \lim_{n\to \infty} h_n(x) .\end{align*}

  • Apply the MVT: \begin{align*} h_n(x) \coloneqq{ h(x, y_n) - h(x, y_0) \over y_n - y_0} &= {\frac{\partial h}{\partial y}\,}(x, \tilde y) && \text{ for some } \tilde y \in [y_0, y_n] .\end{align*}

  • Use this to get a bound for DCT: \begin{align*} {\left\lvert {h_n(x)} \right\rvert} &\coloneqq{\left\lvert { h(x, y_n) - h(x, y_0) \over y_n - y_0} \right\rvert} \\ &= {\left\lvert { {\frac{\partial h}{\partial y}\,}(x, \tilde y) } \right\rvert} \\ &\leq \sup_{y\in [y_0, y_n]} {\left\lvert { {\frac{\partial h}{\partial y}\,}(x, y) } \right\rvert} \\ &\leq \sup_{y\in [y_0, y_n]} {\left\lvert { xf(x) \sin(yx) } \right\rvert} \\ &\leq {\left\lvert { xf(x) } \right\rvert} ,\end{align*} and by assumption \(xf(x) \in L^1\).

  • So this justifies commuting an integral and a limit: \begin{align*} F'(y_0) &\coloneqq\lim_{y_n\to y_0} { F(y_n) - F(y_0) \over y_n - y_0} \\ &= \lim_{n\to 0} \int {h_n(x) } \,dx\\ &\overset{\text{DCT}}{=} \int \lim_{n\to\infty} h_n(x) \,dx\\ &\coloneqq\int {\frac{\partial h}{\partial y}\,}(x, y_0) \,dx\\ &\coloneqq- \int xf(x) \sin(yx) \,dx ,\end{align*} and since this limit exists and is finite, \(F\) is differentiable at \(y_0\).

  • That \(F\) is continuous: \begin{align*} \lim_{y_n \to y_0} F'(y_n) &= \lim_{y_n \to y_0} \int {\frac{\partial h}{\partial y}\,}(x, y_n) \,dx\\ &\overset{\text{DCT}}{=} \int \lim_{y_n \to y_0} {\frac{\partial h}{\partial y}\,}(x, y_n) \,dx\\ &= - \int \lim_{y_n \to y_0} xf(x) \sin(y_n x) \,dx\\ &= - \int xf(x) \sin(y_0x) \,dx ,\end{align*} where we’ve used that \(y\mapsto \sin(yx)\) is clearly continuous.

Spring 2020.3 #real_analysis/qual/completed

  • Prove that if \(g\in L^1({\mathbf{R}})\) then \begin{align*} \lim_{N\to \infty} \int _{{\left\lvert {x} \right\rvert} \geq N} {\left\lvert {f(x)} \right\rvert} \, dx = 0 ,\end{align*} and demonstrate that it is not necessarily the case that \(f(x) \to 0\) as \({\left\lvert {x} \right\rvert}\to \infty\).

  • Prove that if \(f\in L^1([1, \infty])\) and is decreasing, then \(\lim_{x\to\infty}f(x) =0\) and in fact \(\lim_{x\to \infty} xf(x) = 0\).

  • If \(f: [1, \infty) \to [0, \infty)\) is decreasing with \(\lim_{x\to \infty} xf(x) = 0\), does this ensure that \(f\in L^1([1, \infty))\)?

  • Limits
  • Cauchy Criterion for Integrals: \(\int_a^\infty f(x) \,dx\) converges iff for every \({\varepsilon}>0\) there exists an \(M_0\) such that \(A,B\geq M_0\) implies \({\left\lvert {\int_A^B f} \right\rvert} < {\varepsilon}\), i.e. \({\left\lvert {\int_A^B f} \right\rvert} \overset{A\to\infty}\to 0\).
  • Integrals of \(L^1\) functions have vanishing tails: \(\int_{N}^\infty {\left\lvert {f} \right\rvert} \overset{N\to\infty}\longrightarrow 0\).
  • Mean Value Theorem for Integrals: \(\int_a^b f(t)\, dt = (b-a) f(c)\) for some \(c\in [a, b]\).

Stated integral equality:

  • Let \({\varepsilon}> 0\)
  • \(C_c({\mathbf{R}}^n) \hookrightarrow L^1({\mathbf{R}}^n)\) is dense so choose \(\left\{{f_n}\right\} \to f\) with \({\left\lVert {f_n - f} \right\rVert}_1 \to 0\).
  • Since \(\left\{{f_n}\right\}\) are compactly supported, choose \(N_0\gg 1\) such that \(f_n\) is zero outside of \(B_{N_0}(\mathbf{0})\).
  • Then \begin{align*} N\geq N_0 \implies \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f} \right\rvert} &= \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f - f_n + f_n} \right\rvert} \\ &\leq \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f-f_n} \right\rvert} + \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f_n} \right\rvert} \\ &= \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f-f_n} \right\rvert} \\ &\leq \int_{{\left\lvert {x} \right\rvert} > N} {\left\lVert {f-f_n} \right\rVert}_1 \\ &= {\left\lVert {f_n-f} \right\rVert}_1 \qty{\int_{{\left\lvert {x} \right\rvert} > N} 1} \\ &\overset{n\to\infty}\longrightarrow 0 \qty{\int_{{\left\lvert {x} \right\rvert} > N} 1} \\ &= 0\\ &\overset{N\to\infty}\longrightarrow 0 .\end{align*}

To see that this doesn’t force \(f(x)\to 0\) as \({\left\lvert {x} \right\rvert} \to \infty\):

  • Take \(f(x)\) to be a train of rectangles of height 1 and area \(1/2^j\) centered on even integers.
  • Then \begin{align*}\int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f} \right\rvert} = \sum_{j=N}^\infty 1/2^j \overset{N\to\infty}\longrightarrow 0\end{align*} as the tail of a convergent sum.
  • However \(f(x) = 1\) for infinitely many even integers \(x > N\), so \(f(x) \not\to 0\) as \({\left\lvert {x} \right\rvert}\to\infty\).

  • Since \(f\) is decreasing on \([1, \infty)\), for any \(t\in [x-n, x]\) we have \begin{align*} x-n \leq t \leq x \implies f(x) \leq f(t) \leq f(x-n) .\end{align*}

  • Integrate over \([x, 2x]\), using monotonicity of the integral: \begin{align*} \int_x^{2x} f(x) \,dt \leq \int_x^{2x} f(t) \,dt \leq \int_x^{2x} f(x-n) \,dt \\ \implies f(x) \int_x^{2x} \,dt \leq \int_x^{2x} f(t) \,dt \leq f(x-n) \int_x^{2x} \,dt \\ \implies xf(x) \leq \int_x^{2x} f(t) \, dt \leq xf(x-n) .\end{align*}

  • By the Cauchy Criterion for integrals, \(\lim_{x\to \infty} \int_x^{2x} f(t)~dt = 0\).

  • So the LHS term \(xf(x) \overset{x\to\infty}\longrightarrow 0\).

  • Since \(x>1\), \({\left\lvert {f(x)} \right\rvert} \leq {\left\lvert {xf(x)} \right\rvert}\)

  • Thus \(f(x) \overset{x\to\infty}\longrightarrow 0\) as well.

  • Use mean value theorem for integrals: \begin{align*} \int_x^{2x} f(t)\, dt = xf(c_x) \quad\text{for some $c_x \in [x, 2x]$ depending on $x$} .\end{align*}

  • Since \(f\) is decreasing, \begin{align*} x\leq c_x \leq 2x &\implies f(2x)\leq f(c_x) \leq f(x) \\ &\implies 2xf(2x)\leq 2xf(c_x) \leq 2xf(x) \\ &\implies 2xf(2x)\leq 2x\int_x^{2x} f(t)\, dt \leq 2xf(x) \\ .\end{align*}

  • By Cauchy Criterion, \(\int_x^{2x} f \to 0\).

  • So \(2x f(2x) \to 0\), which by a change of variables gives \(uf(u) \to 0\).

  • Since \(u\geq 1\), \(f(u) \leq uf(u)\) so \(f(u) \to 0\) as well.

Just showing \(f(x) \overset{x\to \infty}\longrightarrow 0\):

  • Toward a contradiction, suppose not.

  • Since \(f\) is decreasing, it can not diverge to \(+\infty\)

  • If \(f(x) \to -\infty\), then \(f\not\in L^1({\mathbf{R}})\): choose \(x_0 \gg 1\) so that \(t\geq x_0 \implies f(t) < -1\), then

  • Then \(t\geq x_0 \implies {\left\lvert {f(t)} \right\rvert} \geq 1\), so \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \int_{x_0}^\infty {\left\lvert {f(t) } \right\rvert} \, dt \geq \int_{x_0}^\infty 1 =\infty .\end{align*}

  • Otherwise \(f(x) \to L\neq 0\), some finite limit.

  • If \(L>0\):

    • Fix \({\varepsilon}>0\), choose \(x_0\gg 1\) such that \(t\geq x_0 \implies L-{\varepsilon}\leq f(t) \leq L\)
    • Then \begin{align*}\int_1^\infty f \geq \int_{x_0}^\infty f \geq \int_{x_0}^\infty \qty{L-{\varepsilon}}\,dt = \infty\end{align*}
  • If \(L<0\):

    • Fix \({\varepsilon}> 0\), choose \(x_0\gg 1\) such that \(t\geq x_0 \implies L \leq f(t) \leq L + {\varepsilon}\).
    • Then \begin{align*}\int_1^\infty f \geq \int_{x_0}^\infty f \geq \int_{x_0}^\infty \qty{L}\,dt = \infty\end{align*}

Showing \(xf(x) \overset{x\to \infty}\longrightarrow 0\).

  • Toward a contradiction, suppose not.
  • (How to show that \(xf(x) \not\to + \infty\)?)
  • If \(xf(x)\to -\infty\)
    • Choose a sequence \(\Gamma = \left\{{\widehat{x}_i}\right\}\) such that \(x_i \to \infty\) and \(x_i f(x_i) \to -\infty\).
    • Choose a subsequence \(\Gamma' = \left\{{x_i}\right\}\) such that \(x_if(x_i) \leq -1\) for all \(i\) and \(x_i \leq x_{i+1}\).
    • Choose a further subsequence \(S = \left\{{x_i \in \Gamma' {~\mathrel{\Big\vert}~}2x_i < x_{i+1}}\right\}\).
    • Then since \(f\) is always decreasing, for \(t\geq x_0\), \({\left\lvert {f} \right\rvert}\) is increasing, and \({\left\lvert {f(x_i)} \right\rvert} \leq {\left\lvert {f(2x_i)} \right\rvert}\), so \begin{align*} \int_1^{\infty} {\left\lvert {f} \right\rvert} \geq \int_{x_0}^\infty {\left\lvert {f} \right\rvert} \geq \sum_{x_i \in S} \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert} \, dt \geq \sum_{x_i \in S} \int_{x_i}^{2x_i} {\left\lvert {f(x_i)} \right\rvert} &= \sum_{x_i \in S} x_i f(x_i) \to \infty .\end{align*}
  • If \(xf(x) \to L \neq 0\) for \(0 < L< \infty\):
    • Fix \({\varepsilon}> 0\), choose an infinite sequence \(\left\{{x_i}\right\}\) such that \(L-{\varepsilon}\leq x_i f(x_i) \leq L\) for all \(i\). \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \sum_S \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert}\,dt \geq \sum_S \int_{x_i}^{2x_i} f(x_i) \,dt = \sum_S x_i f(x_i) \geq \sum_S \qty{L-{\varepsilon}} \to \infty .\end{align*}
  • If \(xf(x) \to L \neq 0\) for \(-\infty < L < 0\):
    • Fix \({\varepsilon}> 0\), choose an infinite sequence \(\left\{{x_i}\right\}\) such that \(L \leq x_i f(x_i) \leq L + {\varepsilon}\) for all \(i\). \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \sum_S \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert}\,dt \geq \sum_S \int_{x_i}^{2x_i} f(x_i) \,dt = \sum_S x_i f(x_i) \geq \sum_S \qty{L} \to \infty .\end{align*}

For \(x\geq 1\), \begin{align*} {\left\lvert {xf(x)} \right\rvert} = {\left\lvert { \int_x^{2x} f(x) \, dt } \right\rvert} \leq \int_x^{2x} {\left\lvert {f(x)} \right\rvert} \, dt \leq \int_x^{2x} {\left\lvert {f(t)} \right\rvert}\, dt \leq \int_x^{\infty} {\left\lvert {f(t)} \right\rvert} \,dt \overset{x\to\infty}\longrightarrow 0 \end{align*} where we’ve used

  • Since \(f\) is decreasing and \(\lim_{x\to\infty}f(x) =0\) from part (a), \(f\) is non-negative.
  • Since \(f\) is positive and decreasing, for every \(t\in[a, b]\) we have \({\left\lvert {f(a)} \right\rvert} \leq {\left\lvert {f(t)} \right\rvert}\).
  • By part (a), the last integral goes to zero.

  • Toward a contradiction, produce a sequence \(x_i\to\infty\) with \(x_i f(x_i) \to \infty\) and \(x_if(x_i) > {\varepsilon}> 0\), then \begin{align*} \int f(x) \, dx &\geq \sum_{i=1}^\infty \int_{x_i}^{x_{i+1}} f(x) \, dx \\ &\geq \sum_{i=1}^\infty \int_{x_i}^{x_{i+1}} f(x_{i+1}) \, dx \\ &= \sum_{i=1}^\infty f(x_{i+1}) \int_{x_i}^{x_{i+1}} \, dx \\ &\geq \sum_{i=1}^\infty (x_{i+1} - x_i) f(x_{i+1}) \\ &\geq \sum_{i=1}^\infty (x_{i+1} - x_i) {{\varepsilon}\over x_{i+1}} \\ &= {\varepsilon}\sum_{i=1}^\infty \qty{ 1 - {x_{i-1} \over x_i}} \to \infty \end{align*} which can be ensured by passing to a subsequence where \(\sum {x_{i-1} \over x_i} < \infty\).

  • No: take \(f(x) = {1\over x\ln x}\)
  • Then by a \(u{\hbox{-}}\)substitution, \begin{align*} \int_0^x f = \ln\qty{\ln (x)} \overset{x\to\infty}\longrightarrow\infty \end{align*} is unbounded, so \(f\not\in L^1([1, \infty))\).
  • But \begin{align*} xf(x) = { 1 \over \ln(x)} \overset{x\to\infty} \longrightarrow 0 .\end{align*}

Fall 2019.5 #real_analysis/qual/completed

  • Show that if \(f\) is continuous with compact support on \({\mathbf{R}}\), then \begin{align*} \lim _{y \rightarrow 0} \int_{\mathbb{R}}|f(x-y)-f(x)| d x=0 \end{align*}

  • Let \(f\in L^1({\mathbf{R}})\) and for each \(h > 0\) let \begin{align*} \mathcal{A}_{h} f(x):=\frac{1}{2 h} \int_{|y| \leq h} f(x-y) d y \end{align*}

  • Prove that \(\left\|\mathcal{A}_{h} f\right\|_{1} \leq\|f\|_{1}\) for all \(h > 0\).

  • Prove that \(\mathcal{A}_h f \to f\) in \(L^1({\mathbf{R}})\) as \(h \to 0^+\).

\todo[inline]{Walk through.}

  • Continuity in \(L^1\) (recall that DCT won’t work! Notes 19.4, prove it for a dense subset first).
  • Lebesgue differentiation in 1-dimensional case. See HW 5.6.
  • Fix \(\varepsilon > 0\). If we can find a set \(A\) such that the following calculation holds for \(h\) small enough, we’re done: \begin{align*} \int_{\mathbf{R}}{\left\lvert {f(x-h) - f(x)} \right\rvert} \,dx &= \int_A {\left\lvert {f(x-h) - f(x)} \right\rvert} \,dx\\ &\leq \int_A {\varepsilon}\\ &= {\varepsilon}\mu(A) \longrightarrow 0 ,\end{align*} provided \(h\to 0\) as \({\varepsilon}\to 0\), which we can arrange if \({\left\lvert {h} \right\rvert} < {\varepsilon}\).

  • Choose \(A\supseteq\mathop{\mathrm{supp}}f\) compact such that \(\mathop{\mathrm{supp}}f \pm 1 \subseteq A\)

    • Why this can be done: \(\mathop{\mathrm{supp}}f\) is compact, so closed and bounded, and contained in some compact interval \([-M, M]\). So e.g. \(A\coloneqq[-M-1, M+1]\) suffices.
  • Note that \(f\) is still continuous on \(A\), since it is zero on \(A\setminus\mathop{\mathrm{supp}}f\), and thus uniformly continuous (by Heine-Cantor, for example).

  • We can rephrase the usual definition of uniform continuity: \begin{align*} \forall {\varepsilon}\exists \delta = \delta({\varepsilon}) \text{ such that } {\left\lvert {x - y} \right\rvert} < \delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}\quad \forall x, y\in A \end{align*} as \begin{align*} \forall {\varepsilon}\exists \delta = \delta({\varepsilon}) \text{ such that } {\left\lvert {h} \right\rvert} < \delta \implies {\left\lvert {f(x-h) - f(x)} \right\rvert} < {\varepsilon}\quad \forall x \text{ such that }x, x\pm h \in A \end{align*}

  • So fix \({\varepsilon}\) and choose such a \(\delta\) for \(A\), and choose \(h\) such that \({\left\lvert {h} \right\rvert} < \min(1, \delta)\). Then the desired computation goes through by uniform continuity of \(f\) on \(A\).

We have \begin{align*} \int_{\mathbf{R}}{\left\lvert {A_h(f)(x)} \right\rvert} ~dx &= \int_{\mathbf{R}}{\left\lvert {\frac{1}{2h} \int_{x-h}^{x+h} f(y)~dy} \right\rvert} ~dx \\ &\leq \frac{1}{2h} \int_{\mathbf{R}}\int_{x-h}^{x+h} {\left\lvert {f(y)} \right\rvert} ~dy ~dx \\ &=_{FT} \frac{1}{2h} \int_{\mathbf{R}}\int_{y-h}^{y+h} {\left\lvert {f(y)} \right\rvert} ~\mathbf{dx} ~\mathbf{dy} \\ &= \int_{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} ~{dy} \\ &= {\left\lVert {f} \right\rVert}_1 ,\end{align*}

and (rough sketch)

\begin{align*} \int_{\mathbf{R}}{\left\lvert {A_h(f)(x) - f(x)} \right\rvert} ~dx &= \int_{\mathbf{R}}{\left\lvert { \left(\frac{1}{2h} \int_{B(h, x)} f(y)~dy\right) - f(x)} \right\rvert}~dx \\ &= \int_{\mathbf{R}}{\left\lvert { \left(\frac{1}{2h} \int_{B(h, x)} f(y)~dy\right) - \frac{1}{2h}\int_{B(h, x)} f(x) ~dy} \right\rvert}~dx \\ &\leq_{FT} \frac{1}{2h} \int_{\mathbf{R}}\int_{B(h, x)}{\left\lvert { f(y-x) - f(x)} \right\rvert} ~\mathbf{dx} ~\mathbf{dy} \\ &\leq \frac 1 {2h} \int_{\mathbf{R}}{\left\lVert {\tau_x f - f} \right\rVert}_1 ~dy \\ &\to 0 \quad\text{by (a)} .\end{align*}

This works for arbitrary \(f\in L^1\), using approximation by continuous functions with compact support:

  • Choose \(g\in C_c^0\) such that \({\left\lVert {f- g} \right\rVert}_1 \to 0\).

  • By translation invariance, \({\left\lVert {\tau_h f - \tau_h g} \right\rVert}_1 \to 0\).

  • Write \begin{align*} {\left\lVert {\tau f - f} \right\rVert}_1 &= {\left\lVert {\tau_h f - g + g - \tau_h g + \tau_h g - f} \right\rVert}_1 \\ &\leq {\left\lVert {\tau_h f - \tau_h g} \right\rVert} + {\left\lVert {g - f} \right\rVert} + {\left\lVert {\tau_h g - g} \right\rVert} \\ &\to {\left\lVert {\tau_h g - g} \right\rVert} ,\end{align*}

    so it suffices to show that \({\left\lVert {\tau_h g - g} \right\rVert} \to 0\).

Fall 2017.3 #real_analysis/qual/completed

Let \begin{align*} S = \mathop{\mathrm{span}}_{\mathbf{C}}\left\{{\chi_{(a, b)} {~\mathrel{\Big\vert}~}a, b \in {\mathbf{R}}}\right\}, \end{align*} the complex linear span of characteristic functions of intervals of the form \((a, b)\).

Show that for every \(f\in L^1({\mathbf{R}})\), there exists a sequence of functions \(\left\{{f_n}\right\} \subset S\) such that \begin{align*} \lim _{n \rightarrow \infty}\left\|f_{n}-f\right\|_{1}=0 \end{align*}

  • From homework: \(E\) is Lebesgue measurable iff there exists a finite union of closed cubes \(A\) such that \(m(E\Delta A) < \varepsilon\).

  • Idea: first show this for characteristic functions, then simple functions, then for arbitrary \(f\).

  • For characteristic functions:

    • Consider \(\chi_{A}\) for \(A\) a measurable set.

    • By regularity of the Lebesgue measure, for every \({\varepsilon}>0\) we can find an \(I_{\varepsilon}\) such that \(m(A\Delta I_{\varepsilon})< {\varepsilon}\) where \(I_{\varepsilon}\) is a finite disjoint union of intervals.

    • Then use \begin{align*} {\varepsilon}> m(A\Delta I{\varepsilon}) = \int_X {\left\lvert {\chi_A - \chi_{I_{\varepsilon}}} \right\rvert} ,\end{align*} so the \(\chi_{I_{\varepsilon}}\) converge to \(\chi_A\) in \(L_1\).

    • Then just note that \(\chi_{I_{\varepsilon}} = \sum_{j\leq N} \chi_{I_j}\) where \(I_{\varepsilon}= \coprod_{j\leq N} I_j\), so \(\chi_{I_{\varepsilon}} \in S\).

  • For simple functions:

    • Let \(\psi = \sum_{k\leq N} c_k \chi_{E_k}\).
    • By the argument above, for each \(k\) we can find \(I_{{\varepsilon}, k}\) such that \(\chi_{I_{{\varepsilon}, k}}\) converges to \(\chi_{E_k}\) in \(L^1\).
    • So defining \(\psi_{\varepsilon}= \sum_{k\leq N} c_k \chi_{I_{{\varepsilon}, k}}\), the claim is that this will converge to \(\phi\) in \(L_1\).
    • Note that \begin{align*} \psi_{\varepsilon}= \sum_k c_k \chi_{I_{{\varepsilon}, k}} = \sum_k c_k \sum_j \chi_{I_{j, k} } = \sum_{k, j} c_k \chi_{ I_{j, k} } \in S \end{align*} since now the \(I_{j, k}\) are indicators of intervals.
    • Moreover \begin{align*} {\left\lVert {\psi_{\varepsilon}- \psi} \right\rVert} = {\left\lVert { \sum_k c_k \qty{ \chi_{E_k} - \chi_{I_{{\varepsilon}, k} }} } \right\rVert} \leq \sum_k c_k {\left\lVert { \chi_{E_k} - \chi_{I_{{\varepsilon}, k}} } \right\rVert} ,\end{align*} where the last norm can be bounded by the proof for characteristic functions.
  • For arbitrary functions:

    • Now just use that every \(f \in L^1\) can be approximated by simple functions \(\phi_n\) so that \({\left\lVert {f-\phi_n} \right\rVert}_1 < {\varepsilon}\) for \(n \gg 1\).
    • So find \(\phi_n\to f\), and for each \(n\), find \(g_{n, k} \in S\) with \({\left\lVert {g_{n, k} - \phi_n} \right\rVert}_1 \overset{k\to \infty}\longrightarrow 0\), an approximation by functions in \(S\).
    • Then \begin{align*} {\left\lVert {f - g_{n, k}} \right\rVert} \leq {\left\lVert {f - \phi_n} \right\rVert} + {\left\lVert {\phi_n - g_{n, k}} \right\rVert} ,\end{align*} which can be made arbitrarily small.

Spring 2015.4 #real_analysis/qual/completed

Define \begin{align*} f(x, y):=\left\{\begin{array}{ll}{\frac{x^{1 / 3}}{(1+x y)^{3 / 2}}} & {\text { if } 0 \leq x \leq y} \\ {0} & {\text { otherwise }}\end{array}\right. \end{align*}

Carefully show that \(f \in L^1({\mathbf{R}}^2)\).

Note that

\begin{align*} \int_{{\mathbf{R}}^2}{\left\lvert {f} \right\rvert} \,d\mu &= \int_0^\infty \int_x^\infty x^{1\over 3}(1+xy)^{-3\over 2} \,dy\,dx\\ &= \int_0^\infty -2x^{-{ 2\over 3} }(1+xy)^{-{ 1\over 2} }\Big|_{y=x}^{y=\infty} \,dx\\ &= \int_0^\infty {2\over x^{2\over 3} (1+x^2)^{1\over 2}}\\ &= \int_0^1 {2\over x^{2\over 3} (1+x^2)^{1\over 2}} + \int_1^\infty {2\over x^{2\over 3} (1+x^2)^{1\over 2}} \\ &= \int_0^1 {2\over x^{2\over 3} } + \int_1^\infty {2\over x^{5\over 3} } \\ &<\infty ,\end{align*} where

  • For the first term: We’ve entirely neglected the \(1+x^2\) factor, since neglecting to divide by a positive number can only make the integrand larger,

  • For the second term: \begin{align*} 1+x^2\geq 0 \implies {1\over \sqrt{1+x^2}} \leq {1\over \sqrt{x^2}} = {1\over x} \end{align*}

  • Both terms converge by the \(p{\hbox{-}}\)tests.

The use of iterated integration is justified by Tonelli’s theorem on \({\left\lvert {f} \right\rvert} = f\), since \(f\) is non-negative and clearly measurable on \({\mathbf{R}}^2\), and if any iterated integral is finite then it is equal to \(\int {\left\lvert {f} \right\rvert}\).

Fall 2014.3 #real_analysis/qual/completed

Let \(f\in L^1({\mathbf{R}})\). Show that \begin{align*} \forall\varepsilon > 0 \exists \delta > 0 \text{ such that } \qquad m(E) < \delta \implies \int _{E} |f(x)| \, dx < \varepsilon \end{align*}

  • Note that if \(m(E) = 0\) then \(\int_E f = 0\) for any \(f\).
  • Toward a contradiction, suppose there exists an \({\varepsilon}>0\) such that for all \(\delta>0\) there exists a set \(E_\delta \subseteq {\mathbf{R}}\) with \(m(E) < \delta\) but \(\int_{E_\delta} {\left\lvert {f} \right\rvert} > {\varepsilon}\).
  • Let \(\delta_n \searrow 0\) be any sequence converging to zero and choose \(E_n\) with \(\int_{E_n} {\left\lvert {f} \right\rvert} > {\varepsilon}\) for every \(n\).
  • Define \(E \coloneqq\limsup_n E_n \coloneqq\bigcap_{N\geq 1} \bigcup_{n\geq N} E_n\), then \(m(E) = 0\) by Borel-Cantelli.
  • Now estimate using Fatou: \begin{align*} \int_{E} {\left\lvert {f} \right\rvert} &= \int_X \chi_E {\left\lvert {f} \right\rvert} \\ &= \int_X \limsup_n \chi_{E_n} {\left\lvert {f} \right\rvert} \\ &\geq \limsup_n \int_X \chi_{E_n }{\left\lvert {f} \right\rvert} \\ &\geq \limsup_n \int_{E_n} {\left\lvert {f} \right\rvert} \\ &\geq \limsup_n {\varepsilon}\\ &= {\varepsilon} ,\end{align*} however \(\displaystyle\int_E {\left\lvert {f} \right\rvert}\,dm= 0\) since \(m(E) = 0\), a contradiction. \(\contradiction\).

Note that this is clear for simple functions: let \(\phi = \sum_{k\leq n} c_k m(A_k) < \infty\) be simple function. then \(\phi\) is necessarily bounded on \({\mathbf{R}}\), so let \(M\coloneqq\sup_{\mathbf{R}}\phi\) and estimate \begin{align*} \int_E \phi &\coloneqq\sum_k c_k m(A_k \cap E) \\ &\leq \sum_k M\cdot m(E)\\ &= C M m(E) ,\end{align*} for some constant \(C\), so choosing \(\delta < { {\varepsilon}\over C M}\) (and its corresponding \(E\) with \(m(E) < \delta\)) bounds this above by \({\varepsilon}\).

For arbitrary \(f \in L^1\), there is a sequence of simple functions \(\phi_n\) with \(\int \phi_n \nearrow\int f\) and \({\left\lVert {\phi_n - f} \right\rVert}_{L_1} \overset{n\to\infty}\longrightarrow 0\). Choose \(\delta\) and \(E\) as above, and use the triangle inequality to estimate \begin{align*} \int_E {\left\lvert {f} \right\rvert} &= \int_E {\left\lvert {f - \phi_n + \phi_n} \right\rvert} \\ &\leq \int_E {\left\lvert {f - \phi_n} \right\rvert} + \int_E {\left\lvert {\phi_n} \right\rvert} ,\end{align*} choose \(n\gg 1\) to bound the first term by \({\varepsilon}\), noting that the second term is bounded by \({\varepsilon}\) by the case for simple functions.

Spring 2014.1 #real_analysis/qual/completed

  • Give an example of a continuous \(f\in L^1({\mathbf{R}})\) such that \(f(x) \not\to 0\) as\({\left\lvert {x} \right\rvert} \to \infty\).

  • Show that if \(f\) is uniformly continuous, then \begin{align*} \lim_{{\left\lvert {x} \right\rvert} \to \infty} f(x) = 0. \end{align*}

Part 1: Take a train of triangles with base points at \(k\) and \(k+1\), each of area \(2^{-k}\). Then \(\int {\left\lvert {f} \right\rvert} \approx \sum_{k\geq 0} 2^{-k} <\infty\), but \(f(x)\not\to 0\) since \(f(x) > 0\) infinitely often.

Part 2:

  • Idea: use contradiction to produce a sequence with arbitrarily large terms, and bound below an integral in a ball about each point.

  • Suppose \(\lim_{{\left\lvert {x} \right\rvert}\to \infty}f(x) = L > 0\).

    • Then for any \({\varepsilon}\) there exists an \(M\) such that \(x\geq M \implies {\left\lvert {f(x) - L} \right\rvert} < {\varepsilon}\), so \(L-{\varepsilon}\leq f(x) \leq L+{\varepsilon}\)
    • Choosing \({\varepsilon}=L/2\) yields \(L/2 \leq f(x) \leq 3L/2\), and so \begin{align*} \int_{\mathbf{R}}{\left\lvert {f} \right\rvert} \geq \int_{{\left\lvert {x} \right\rvert} \geq M} {\left\lvert {f} \right\rvert} \geq \int_{{\left\lvert {x} \right\rvert}\geq M} L/2 \to \infty ,\end{align*} contradicting \(f\in L^1({\mathbf{R}})\). \(\contradiction\).
  • So it must be that this limit does not exist. Fix \({\varepsilon}>0\), then there are infinitely many \(x\) such that \(f(x) > {\varepsilon}\), so choose a sequence \(x_n\to \infty\) with \(f(x_n) > {\varepsilon}\) for each \(n\).

  • Now use uniform continuity: pick a uniform \(\delta = \delta({\varepsilon})\) such that \(x\in B_\delta(x_n) \implies {\left\lvert {f(x) - f(x_n)} \right\rvert} < {\varepsilon}/4\).

  • Now use that \(f(x_n) - {\varepsilon}/4 \leq f(x) \leq f(x_n)+{\varepsilon}/4\) implies that \(f(x) \geq 3{\varepsilon}/4\) whenever \(x\in B_\delta(x_n)\) for any \(n\) to estimate \begin{align*} \int_{B_\delta(x_n)} {\left\lvert {f(x)} \right\rvert}\,dx \geq 2\delta \cdot 3{\varepsilon}/4 \coloneqq C = C_{\delta, {\varepsilon}} > 0 ,\end{align*} where \(C\) is a constant.

  • But now we’ve contradicted \(f\in L^1\): \begin{align*} \int_{\mathbf{R}}{\left\lvert {f} \right\rvert} \geq \sum_{n\geq 1} \int_{B_\delta(x_n)} {\left\lvert {f} \right\rvert} \geq \sum_{n\geq 1} C \to \infty ,\end{align*} provided we pass to a further subsequence of \(x_n\) such that the balls \(B_\delta(x_n)\) are disjoint. \(\contradiction\)

#real_analysis/qual/completed #todo #real_analysis/qual/work