\(L^2\) and Fourier Analysis

Fall 2020.5 #real_analysis/qual/completed

Suppose \(\varphi\in L^1({\mathbf{R}})\) with \begin{align*} \int \varphi(x) \, dx = \alpha .\end{align*} For each \(\delta > 0\) and \(f\in L^1({\mathbf{R}})\), define \begin{align*} A_\delta f(x) \coloneqq\int f(x-y) \delta^{-1} \varphi\qty{\delta^{-1} y}\, dy .\end{align*}

  • Prove that for all \(\delta > 0\), \begin{align*} {\left\lVert {A_\delta f} \right\rVert}_1 \leq {\left\lVert {\varphi} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 .\end{align*}

  • Prove that \begin{align*} A_\delta f \to \alpha f \text{ in } L^1({\mathbf{R}}) {\quad \operatorname{as} \quad} \delta\to 0^+ .\end{align*}

Hint: you may use without proof the fact that for all \(f\in L^1({\mathbf{R}})\), \begin{align*} \lim_{y\to 0} \int_{\mathbf{R}}{\left\lvert {f(x-y) - f(x)} \right\rvert}\, dx = 0 .\end{align*}

See Folland 8.14.

This is a direct application of Fubini-Tonelli: \begin{align*} {\left\lVert {A_\delta f} \right\rVert} &\coloneqq\int {\left\lvert { \int f(x-y)\delta^{-1}\varphi(\delta^{-1}y)\,dy} \right\rvert} \,dx\\ &\leq \int \int {\left\lvert {f(x-y)\delta^{-1}\varphi(\delta^{-1}y)} \right\rvert} \,dy\,dx\\ &\overset{FT}{=} \int \int {\left\lvert { f(x-y) } \right\rvert} \cdot {\left\lvert {\delta^{-1}\varphi(\delta^{-1}y)} \right\rvert} \,dx\,dy\\ &= \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y) } \right\rvert} \qty{ \int {\left\lvert { f(x-y) } \right\rvert}\,dx} \,dy\\ &= \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y)} \right\rvert}\cdot {\left\lVert {f} \right\rVert} \,dy\\ &= {\left\lVert {f} \right\rVert} \cdot \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y) } \right\rvert} \,dy\\ &= {\left\lVert {f} \right\rVert} \cdot {\left\lVert {\varphi} \right\rVert} .\end{align*} Here we’ve used translation and dilation invariance of the Lebesgue integral.

Write \(\phi_\delta(y) \coloneqq\delta^{-1}\phi(\delta^{-1}y)\), then \begin{align*} {\left\lVert {A_\delta f - \alpha f} \right\rVert}_1 &\coloneqq\int {\left\lvert {A_\delta f(x) - \alpha f(x) } \right\rvert} \,dx\\ &= \int {\left\lvert { \int {f(x-y) \phi_\delta(y) } \,dy- \alpha f(x) } \right\rvert}\,dx\\ &= \int {\left\lvert { \int { \tau_y f (x) \phi_\delta(y) } \,dy- \int f(x) \phi_\delta(y) \,dy} \right\rvert}\,dx\\ &\leq \int\int {\left\lvert {\tau_y f(x) - f(x)} \right\rvert}\cdot {\left\lvert {\phi_\delta(y)} \right\rvert} \,dy\,dx\\ &= \int\int {\left\lvert {\tau_y f(x) - f(x)} \right\rvert}\cdot {\left\lvert {\phi_\delta(y)} \right\rvert} \,dx\,dy\\ &= \int{\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}_1 \,dy ,\end{align*} where the interchange of integration order is justified by Tonelli since the integrands are positive. The goal is to now make this small when \(\delta\) is small.

One way to do this immediately: make a change of variables \(y=tz\) to get \begin{align*} {\left\lVert {A_\delta f - \alpha f} \right\rVert}_1 \leq\int {{\left\lvert {\phi(z)} \right\rvert}} {\left\lVert {\tau_{tz}f -f} \right\rVert}_1 \,dz ,\end{align*} use that \({\left\lVert {\tau_{tz} f- f} \right\rVert}_1 \leq 2{\left\lVert {f} \right\rVert}_1 < \infty\) by the triangle inequality and apply the DCT: \begin{align*} \lim_{t\to 0} \int {{\left\lvert {\phi(z)} \right\rvert}} \cdot {\left\lVert {\tau_{tz}f -f} \right\rVert}_1 \,dz= \int {{\left\lvert {\phi(z)} \right\rvert}} \lim_{t\to 0} {\left\lVert {\tau_{tz}f -f} \right\rVert}_1 \,dz= 0 .\end{align*}

More directly, use continuity in \(L^1\) (as per the hint) to pick a \(h>0\) such that \begin{align*} {\left\lVert {\tau_y f - f} \right\rVert}< {\varepsilon}\quad \text{ for } y\in A \coloneqq\left\{{y{~\mathrel{\Big\vert}~}{\left\lvert {y} \right\rvert} \leq h}\right\} .\end{align*} Now choose \(\delta_0 \gg 1\) large enough so that \begin{align*} \int_{A^c}{\left\lvert {\phi_\delta(y)} \right\rvert}\,dy< {\varepsilon}\quad \text{ for all }\delta > \delta_0 .\end{align*} Now \begin{align*} \int_{\mathbf{R}}{\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}_1 \,dy &= \int_A {\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}_1 \,dy+ \int_{A^c} {\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\left\lVert {\tau_yf - f} \right\rVert}_1 \,dy\\ &\leq \int_A {\left\lvert {\phi_\delta(y)} \right\rvert}\cdot {\varepsilon}\,dy+ \int_{A^c} {\left\lvert {\phi_\delta(y)} \right\rvert} \cdot 2{\left\lVert {f} \right\rVert}_1 \,dy\\ &\leq {\varepsilon}{\left\lVert {\phi_\delta} \right\rVert}_1 + 2{\varepsilon}{\left\lVert {f} \right\rVert}_1 \\ &\longrightarrow 0 .\end{align*}

Spring 2020.6 #real_analysis/qual/completed

  • Show that \begin{align*} L^2([0, 1]) \subseteq L^1([0, 1]) {\quad \operatorname{and} \quad} \ell^1({\mathbf{Z}}) \subseteq \ell^2({\mathbf{Z}}) .\end{align*}

  • For \(f\in L^1([0, 1])\) define \begin{align*} \widehat{f}(n) \coloneqq\int _0^1 f(x) e^{-2\pi i n x} \, dx .\end{align*} Prove that if \(f\in L^1([0, 1])\) and \(\left\{{\widehat{f}(n)}\right\} \in \ell^1({\mathbf{Z}})\) then \begin{align*} S_N f(x) \coloneqq\sum_{{\left\lvert {n} \right\rvert} \leq N} \widehat{f} (n) e^{2 \pi i n x} .\end{align*} converges uniformly on \([0, 1]\) to a continuous function \(g\) such that \(g = f\) almost everywhere.

Hint: One approach is to argue that if \(f\in L^1([0, 1])\) with \(\left\{{\widehat{f} (n)}\right\} \in \ell^1({\mathbf{Z}})\) then \(f\in L^2([0, 1])\).

From Neil:

  • \(\widehat{f}\) in \(\ell^1\) ensures that \(S_N\) converges uniformly to something, call it \(g\).
  • \(\widehat{f} \in\ell^1\) Implies \(\widehat{f}\in \ell^2\) which (by characterization of an o.n.b.) implies \(f\) is in \(L^2\) (Parseval) and (again by characterization of an o.n.b.) that \(S_N\) converges to \(f\) in \(L^2\) (and hence a subsequence converges to f a.e.)
  • By uniqueness of limits \(f=g\).

Other stuff:

  • For \(e_n(x) \coloneqq e^{2\pi i n x}\), the set \(\left\{{e_n}\right\}\) is an orthonormal basis for \(L^2([0, 1])\).
  • For any orthonormal sequence in a Hilbert space, we have Bessel’s inequality: \begin{align*} \sum_{k=1}^{\infty}\left|\left\langle x, e_{k}\right\rangle\right|^{2} \leq\|x\|^{2} .\end{align*}
  • When \(\left\{{e_n}\right\}\) is a basis, the above is an equality (Parseval)
  • Arguing uniform convergence: since \(\left\{{\widehat{f}(n)}\right\} \in \ell^1({\mathbf{Z}})\), we should be able to apply the \(M\) test.

Claim: if \(f\in L^1[0, 1]\) and \(\widehat{f}\in \ell^1({\mathbf{Z}})\), then \(S_Nf \to f\) uniformly.

  • Since \(\widehat{f}\in \ell^1({\mathbf{Z}})\), we have \(S_Nf\to g\) uniformly for some continuous \(g\) by the \(M{\hbox{-}}\)test.

  • Now consider \(\widehat{g}\). We have \begin{align*} \widehat{g}(n) = \int_0^1 \sum_m \qty{\widehat{f}(m)e_m(x)}e_{-n}(x) \,dx= \widehat{f}(n) ,\end{align*} using that \(\int_0^1 e_n(x)\,dx= \chi_{n=0}\).

  • We’ll now show \(f-g= 0\) a.e. by mollifying against an approximate identity \(\varphi\in L^1\), setting \begin{align*} \varphi_{\varepsilon}(x) \coloneqq{\varepsilon}^{-1}\varphi({\varepsilon}^{-1}x) \in L^1[0, 1] .\end{align*}

  • A computation: \begin{align*} \widehat{f\ast\varphi_{\varepsilon}}(n) &= \widehat{f}\cdot \widehat{\varphi_{\varepsilon}}(n) \\ &= \widehat{g}\cdot \widehat{\varphi_{\varepsilon}}(n) \\ &= \widehat{g\ast\varphi_{\varepsilon}}(n) ,\end{align*} so \begin{align*} \widehat{(f-g)\ast\varphi_{\varepsilon}} = 0 \quad \forall n \implies (f-g)\ast\varphi_{\varepsilon}\equiv 0 ,\end{align*} using that \((f-g)\ast\varphi_{\varepsilon}\in L^2\) and \(\left\{{e_n}\right\}\) for a complete orthonormal basis of \(L^2\).

  • Now use that \((f-g)\ast\varphi_{\varepsilon}\to (f-g)\) in \(L^1\) and \((f-g)\ast\varphi_{\varepsilon}\equiv 0\) to conclude \(f-g = 0\) a.e.

\(\ell^1({\mathbf{Z}}) \subseteq \ell^2({\mathbf{Z}})\).

  • Set \(\mathbf{c} = \left\{{c_k {~\mathrel{\Big\vert}~}k\in {\mathbf{Z}}}\right\} \in \ell^1({\mathbf{Z}})\).
  • It suffices to show that if \(\sum_{k\in {\mathbf{Z}}} {\left\lvert {c_k} \right\rvert} < \infty\) then \(\sum_{k\in {\mathbf{Z}}} {\left\lvert {c_k} \right\rvert}^2 < \infty\).
  • Let \(S = \left\{{c_k {~\mathrel{\Big\vert}~}{\left\lvert {c_k} \right\rvert} \leq 1}\right\}\), then \(c_k \in S \implies {\left\lvert {c_k} \right\rvert}^2 \leq {\left\lvert {c_k} \right\rvert}\)
  • Claim: \(S^c\) can only contain finitely many elements, all of which are finite.
    • If not, either \(S^c \coloneqq\left\{{c_j}\right\}_{j=1}^\infty\) is infinite with every \({\left\lvert {c_j} \right\rvert} > 1\), which forces \begin{align*}\sum_{c_k\in S^c} {\left\lvert {c_k} \right\rvert} = \sum_{j=1}^\infty {\left\lvert {c_j} \right\rvert} > \sum_{j=1}^\infty 1 = \infty.\end{align*}
    • If any \(c_j = \infty\), then \(\sum_{k\in {\mathbf{Z}}} {\left\lvert {c_k} \right\rvert} \geq c_j = \infty\).
  • So \(S^c\) is a finite set of finite integers, let \(N = \max \left\{{{\left\lvert {c_j} \right\rvert}^2 {~\mathrel{\Big\vert}~}c_j \in S^c}\right\} < \infty\).
  • Rewrite the sum \begin{align*} \sum_{k\in {\mathbf{Z}}} {\left\lvert {c_k} \right\rvert}^2 &= \sum_{c_k\in S} {\left\lvert {c_k} \right\rvert}^2 + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq \sum_{c_k\in S} {\left\lvert {c_k} \right\rvert} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq \sum_{k\in {\mathbf{Z}}} {\left\lvert {c_k} \right\rvert} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \quad\text{since the ${\left\lvert {c_k} \right\rvert}$ are all positive} \\ &= {\left\lVert {\mathbf{c}} \right\rVert}_{\ell^1} + \sum_{c_k \in S^c} {\left\lvert {c_k} \right\rvert}^2 \\ &\leq {\left\lVert {\mathbf{c}} \right\rVert}_{\ell^1} + {\left\lvert {S^c} \right\rvert}\cdot N \\ &< \infty .\end{align*}

\(L^2([0, 1]) \subseteq L^1([0, 1])\).

  • It suffices to show that \(\int {\left\lvert {f} \right\rvert}^2 < \infty \implies \int {\left\lvert {f} \right\rvert} < \infty\).

  • Define \(S = \left\{{x\in [0, 1] {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} \leq 1}\right\}\), then \(x\in S^c \implies {\left\lvert {f(x)} \right\rvert}^2 \geq {\left\lvert {f(x)} \right\rvert}\).

  • Break up the integral: \begin{align*} \int_{\mathbf{R}}{\left\lvert {f} \right\rvert} &= \int_S {\left\lvert {f} \right\rvert} + \int_{S^c} {\left\lvert {f} \right\rvert} \\ &\leq \int_S {\left\lvert {f} \right\rvert} + \int_{S^c} {\left\lvert {f} \right\rvert}^2 \\ &\leq \int_S {\left\lvert {f} \right\rvert} + {\left\lVert {f} \right\rVert}_2 \\ &\leq \sup_{x\in S}\left\{{{\left\lvert {f(x)} \right\rvert}}\right\} \cdot \mu(S) + {\left\lVert {f} \right\rVert}_2 \\ &= 1 \cdot \mu(S) + {\left\lVert {f} \right\rVert}_2 \quad\text{by definition of } S\\ &\leq 1 \cdot \mu([0, 1]) + {\left\lVert {f} \right\rVert}_2 \quad\text{since } S\subseteq [0, 1] \\ &= 1 + {\left\lVert {f} \right\rVert}_2 \\ &< \infty .\end{align*}

Note: this proof shows \(L^2(X) \subseteq L^1(X)\) whenever \(\mu(X) < \infty\).

  • First, \(S_Nf\) converges in \({\mathcal{H}}\) to something, say \(g \coloneqq\lim_{n\to\infty} S_n f\), since \begin{align*} {\left\lVert {g - S_Nf} \right\rVert} = {\left\lVert {\sum_{{\left\lvert {n} \right\rvert} \geq N} \widehat{f} (n) e_n(x) } \right\rVert} \leq \sum_{{\left\lvert {n} \right\rvert} \geq N } {\left\lvert {\widehat{f}(n)} \right\rvert} \overset{N\to\infty}\longrightarrow 0 ,\end{align*} where the last term is the tail of a convergent sum since \(\left\{{\widehat{f}(n)}\right\} \in \ell^1\).

  • This also shows that \(S_N\to g\) uniformly.

  • \(g\) is continuous, as the uniform limit of continuous functions.

  • Showing that \(g = f\) a.e.: it suffices to show that \(S_N\) converges to \(f\) in \(L^p\) for some \(p\), since this will provide a subsequence that converges to \(f\) a.e..

  • Claim: \(\widehat{f}\in \ell^1 \subseteq \ell^2\) implies that \(f \in L^2\). This follows from Parseval: \begin{align*} \infty > {\left\lVert {\widehat{f}} \right\rVert}_{\ell^2}^2 = \sum_{n\in {\mathbf{Z}}} {\left\lvert {\widehat{f}(n)} \right\rvert}^2 = \int_0^1 {\left\lvert {f(z)} \right\rvert}^2 \,dz = {\left\lVert {f} \right\rVert}_{L^2}^2 .\end{align*}

  • Claim: \(S_N\to f\) in \(L^2\).

    • This follows from the fact that \(\left\{{e_n}\right\}_{n\in {\mathbf{Z}}}\) is a complete orthonormal basis, so \(f = \sum {\left\langle {f},~{e_n} \right\rangle}e_n\) uniquely, recognizing \(\widehat{f}(n) = {\left\langle {f},~{e_n} \right\rangle}\), and writing \begin{align*} f = \sum_{n} {\left\langle {f},~{e_n} \right\rangle}e_n = \sum_n \widehat{f}(n) e_n \coloneqq\lim_{N\to\infty }S_N f .\end{align*}
  • So a subsequence \(\left\{{S_{N_k}}\right\}_{k\geq 0}\) converges to \(f\) a.e.. Since \(S_N\to g\) a.e., \(f=g\) a.e. by uniqueness of limits.

Fall 2017.5 #real_analysis/qual/completed

Let \(\varphi\) be a compactly supported smooth function that vanishes outside of an interval \([-N, N]\) such that \(\int _{{\mathbf{R}}} \varphi(x) \, dx = 1\).

For \(f\in L^1({\mathbf{R}})\), define \begin{align*} K_{j}(x) \coloneqq j \varphi(j x), \qquad f \ast K_{j}(x) \coloneqq\int_{{\mathbf{R}}} f(x-y) K_{j}(y) \, dy \end{align*} and prove the following:

  • Each \(f\ast K_j\) is smooth and compactly supported.

  • \begin{align*} \lim _{j \to \infty} {\left\lVert {f * K_{j}-f} \right\rVert}_{1} = 0 \end{align*}

Hint: \begin{align*} \lim _{y \to 0} \int _{{\mathbf{R}}} |f(x-y)-f(x)| dy = 0 \end{align*}

\todo[inline]{Add concepts.}
  • ?

Part a

Lemma: If \(\varphi \in C_c^1\), then \((f \ast \varphi)' = f \ast \varphi'\) almost everywhere.

Silly Proof:

\begin{align*} \mathcal{F}( (f \ast \varphi)' ) &= 2\pi i \xi ~\mathcal{F}(f\ast \varphi) \\ &= 2\pi i \xi ~ \mathcal{F}(f) ~ \mathcal{F}(\varphi) \\ &= \mathcal{F}(f) \cdot \left( 2\pi i \xi ~\mathcal{F}(\varphi)\right) \\ &= \mathcal{F}(f) \cdot \mathcal{F}(\varphi') \\ &= \mathcal{F}(f\ast \varphi') .\end{align*}

Actual proof:

\begin{align*} (f\ast \varphi)'(x) &= (\varphi\ast f)'(x) \\ &= \lim_{h\to 0} \frac{(\varphi\ast f)'(x+h) - (\varphi\ast f)'(x)}{h} \\ &= \lim_{h\to 0} \int \frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y) \\ &\overset{DCT}= \int \lim_{h\to 0} \frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y) \\ &= \int \varphi'(x-y) f(y) \\ &= (\varphi' \ast f)(x) \\ &= (f \ast \varphi')(x) .\end{align*}

To see that the DCT is justified, we can apply the MVT on the interval \([0, h]\) to \(f\) to obtain

\begin{align*} \frac{\varphi(x + h - y) - \varphi(x - y)}{h} &= \varphi'(c) \quad c\in [0, h] ,\end{align*}

and since \(\varphi'\) is continuous and compactly supported, \(\varphi'\) is bounded by some \(M < \infty\) by the extreme value theorem and thus \begin{align*} \int {\left\lvert {\frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y)} \right\rvert} &= \int {\left\lvert {\varphi'(c) f(y)} \right\rvert} \\ &\leq \int {\left\lvert {M} \right\rvert}{\left\lvert {f} \right\rvert} \\ &= {\left\lvert {M} \right\rvert} \int {\left\lvert {f} \right\rvert} < \infty ,\end{align*}

since \(f\in L^1\) by assumption, so we can take \(g\coloneqq{\left\lvert {M} \right\rvert} {\left\lvert {f} \right\rvert}\) as the dominating function.

Applying this theorem infinitely many times shows that \(f\ast \varphi\) is smooth.

To see that \(f\ast \varphi\) is compactly supported, approximate \(f\) by a continuous compactly supported function \(h\), so \({\left\lVert {h - f} \right\rVert}_1 \overset{L^1}\to 0\).

Now let \(g_x(y) = \varphi(x-y)\), and note that \(\mathrm{supp}(g) = x - \mathrm{supp}(\varphi)\) which is still compact.

But since \(\mathrm{supp}(h)\) is bounded, there is some \(N\) such that \begin{align*} {\left\lvert {x} \right\rvert} > N \implies A_x\coloneqq\mathrm{supp}(h) \cap\mathrm{supp}(g_x) = \emptyset \end{align*}

and thus \begin{align*} (h\ast \varphi)(x) &= \int_{\mathbf{R}}\varphi(x-y) h(y)~dy \\ &= \int_{A_x} g_x(y) h(y) \\ &= 0 ,\end{align*}

so \(\left\{{x {~\mathrel{\Big\vert}~}f\ast g(x) = 0}\right\}\) is open, and its complement is closed and bounded and thus compact.

Part b

\begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - f(x)} \right\rvert}~dx \\ &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - \int f(x) K_j(y) ~ dy} \right\rvert}~dx \\ &= \int {\left\lvert {\int ( f(x-y) - f(x) ) K_j(y) ~dy } \right\rvert} ~dx \\ &\leq \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} ~ dy~dx \\ &\overset{FT}= \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} \mathbf{~ dx~dy}\\ &= \int {\left\lvert {K_j(y)} \right\rvert} \left( \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} ~ dx\right) ~dy \\ &= \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy .\end{align*}

We now split the integral up into pieces.

  • Chose \(\delta\) small enough such that \({\left\lvert {y} \right\rvert} < \delta \implies {\left\lVert {f - \tau_y f} \right\rVert}_1 < \varepsilon\) by continuity of translation in \(L^1\), and

  • Since \(\varphi\) is compactly supported, choose \(J\) large enough such that \begin{align*} j > J \implies \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} ~dy = \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {j\varphi(jy)} \right\rvert} = 0 \end{align*}

Then \begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &\leq \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy \\ &= \int_{{\left\lvert {y} \right\rvert} < \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy+ \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy \\ &= \varepsilon \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} + 0 \\ &\leq \varepsilon(1) \to 0 .\end{align*}

Spring 2017.5 #real_analysis/qual/work

Let \(f, g \in L^2({\mathbf{R}})\). Prove that the formula \begin{align*} h(x) \coloneqq\int _{-\infty}^{\infty} f(t) g(x-t) \, dt \end{align*} defines a uniformly continuous function \(h\) on \({\mathbf{R}}\).

Spring 2015.6 #real_analysis/qual/work

Let \(f \in L^1({\mathbf{R}})\) and \(g\) be a bounded measurable function on \({\mathbf{R}}\).

  • Show that the convolution \(f\ast g\) is well-defined, bounded, and uniformly continuous on \({\mathbf{R}}\).
  • Prove that one further assumes that \(g \in C^1({\mathbf{R}})\) with bounded derivative, then \(f\ast g \in C^1({\mathbf{R}})\) and \begin{align*} \frac{d}{d x}(f * g)=f *\left(\frac{d}{d x} g\right) \end{align*}

Fall 2014.5 #real_analysis/qual/work

  • Let \(f \in C_c^0({\mathbf{R}}^n)\), and show \begin{align*} \lim _{t \to 0} \int_{{\mathbf{R}}^n} |f(x+t) - f(x)| \, dx = 0 .\end{align*}

  • Extend the above result to \(f\in L^1({\mathbf{R}}^n)\) and show that \begin{align*} f\in L^1({\mathbf{R}}^n), \quad g\in L^\infty({\mathbf{R}}^n) \quad \implies f \ast g \text{ is bounded and uniformly continuous. } \end{align*}

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