Functional Analysis: General

Fall 2019.4 #real_analysis/qual/completed

Let \(\{u_n\}_{n=1}^∞\) be an orthonormal sequence in a Hilbert space \(\mathcal{H}\).

  • Prove that for every \(x ∈ \mathcal H\) one has \begin{align*} \displaystyle\sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|^{2} \end{align*}

  • Prove that for any sequence \(\{a_n\}_{n=1}^\infty \in \ell^2({\mathbb{N}})\) there exists an element \(x\in\mathcal H\) such that \begin{align*} a_n = {\left\langle {x},~{u_n} \right\rangle} \text{ for all } n\in {\mathbb{N}} \end{align*} and \begin{align*} {\left\lVert {x} \right\rVert}^2 = \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \end{align*}

  • Bessel’s Inequality
  • Pythagoras
  • Surjectivity of the Riesz map
  • Parseval’s Identity
  • Trick – remember to write out finite sum \(S_N\), and consider \({\left\lVert {x - S_N} \right\rVert}\).

  • Equivalently, we can show \begin{align*} {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^\infty {\left\lvert { {\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \geq 0 .\end{align*}

  • Claim: the LHS is the norm of an element in \(H\), and thus non-negative. More precisely, set \(S_N\coloneqq\sum_{n=1}^N {\left\langle {x},~{u_n} \right\rangle}u_n\), then the above is equal to \begin{align*} {\left\lVert {x - \lim_{N\to\infty} S_N} \right\rVert}^2 .\end{align*} Note that if this is true, we’re done.

  • To see this, expand the norm in terms of inner products: ` \begin{align*} {\left\lVert {x - S_N} \right\rVert}^2 &= {\left\langle {x-S_N},~{x-S_N} \right\rangle} \ &= {\left\langle {x},~{x} \right\rangle} - {\left\langle {x},~{S_N} \right\rangle} - {\left\langle {S_N},~{x} \right\rangle} + {\left\langle {S_N},~{S_N} \right\rangle} \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - \qty{{\left\langle {x},~{S_N} \right\rangle} + {\overline{{{\left\langle {x},~{S_N} \right\rangle}}}} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{{\left\langle {x},~{S_N} \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{ {\left\langle {x},~{\sum_^N {\left\langle {x},~{u_n} \right\rangle} u_n } \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{ \sum_^N {\left\langle {x},~{{\left\langle {x},~{u_n} \right\rangle} u_n } \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re\qty{ \sum_^N {\overline{{{\left\langle {x},~{u_n} \right\rangle} }}} {\left\langle {x},~{u_n } \right\rangle} } \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\Re \sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_N} \right\rVert}^2 - 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {\sum_^N {\left\langle {x},~{u_n} \right\rangle} u_n} \right\rVert}^2 - 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + {\left\langle {\sum_^N {\left\langle {x},~{u_n} \right\rangle} u_n},~{\sum_^N {\left\langle {x},~{u_m} \right\rangle} u_m} \right\rangle}

    • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + \sum_{n, m \leq N}{\left\langle {x},~{u_n} \right\rangle} {\overline{{{\left\langle {x},~{u_m} \right\rangle} }}}{\left\langle {u_n},~{u_m} \right\rangle}
    • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + \sum_{n, m\leq N} {\left\langle {x},~{u_n} \right\rangle} {\overline{{{\left\langle {x},~{u_m} \right\rangle}}}} \delta_{mn}
    • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2 + \sum_{n\leq N} {\left\lvert {{\left\langle {x},~{u_n} \right\rangle}} \right\rvert}^2
    • 2\sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 \ &= {\left\lVert {x} \right\rVert}^2
    • \sum_^N {\left\lvert {{\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 .\end{align*} `{=html}
  • Now take \(\lim_{N\to\infty}\) and use that \({\left\lVert {{-}} \right\rVert}\) is continuous.

  • Set \begin{align*} x\coloneqq\sum_{n\in {\mathbb{N}}} a_n u_n .\end{align*}

  • Checking the first desired property: \begin{align*} {\left\langle {x},~{u_m} \right\rangle} &= {\left\langle { \sum_{n\geq 1} a_n u_n },~{u_m} \right\rangle} \\ &=\sum_{n\geq 1} a_n {\left\langle { u_n },~{u_m} \right\rangle} \\ &=\sum_{n\geq 1} a_n \delta_{mn} \\ &= a_m .\end{align*}

  • That \(x\in H\): this would follow from \begin{align*} {\left\lVert {x} \right\rVert}^2 = \sum_n {\left\lvert {{\left\langle {x},~{u_n } \right\rangle}} \right\rvert}^2 = \sum_n {\left\lvert {a_n} \right\rvert}^2 <\infty .\end{align*} The inequality holds by assumption since \(\left\{{a_n}\right\}\in\ell^2\), so it suffices to show the first equality:

\begin{align*} {\left\lVert {x} \right\rVert}^2 &\coloneqq{\left\langle {x},~{x} \right\rangle} \\ &= {\left\langle {\sum_n a_n u_n},~{\sum_m a_m u_m} \right\rangle} \\ &= \sum_{n, m} a_n {\overline{{a_m}}} {\left\langle {u_n},~{u_m} \right\rangle} \\ &= \sum_{n, m} a_n {\overline{{a_m}}} \delta_{mn} \\ &= \sum_{n} a_n {\overline{{a_n}}} \\ &= \sum_{n} {\left\lvert {a_n} \right\rvert}^2 \\ &= \sum_n {\left\lvert {{\left\langle {x},~{u_n} \right\rangle}} \right\rvert}^2 .\end{align*}

Spring 2019.5 #real_analysis/qual/completed

  • Show that \(L^2([0, 1]) ⊆ L^1([0, 1])\) and argue that \(L^2([0, 1])\) in fact forms a dense subset of \(L^1([0, 1])\).

  • Let \(Λ\) be a continuous linear functional on \(L^1([0, 1])\).

Prove the Riesz Representation Theorem for \(L^1([0, 1])\) by following the steps below:

  • Establish the existence of a function \(g ∈ L^2([0, 1])\) which represents \(Λ\) in the sense that \begin{align*} Λ(f ) = f (x)g(x) dx \text{ for all } f ∈ L^2([0, 1]). \end{align*}

Hint: You may use, without proof, the Riesz Representation Theorem for \(L^2([0, 1])\).

  • Argue that the \(g\) obtained above must in fact belong to \(L^∞([0, 1])\) and represent \(Λ\) in the sense that \begin{align*} \Lambda(f)=\int_{0}^{1} f(x) \overline{g(x)} d x \quad \text { for all } f \in L^{1}([0,1]) \end{align*} with \begin{align*} \|g\|_{L^{\infty}([0,1])} = \|\Lambda\|_{L^{1}([0,1]) {}^{ \vee }} \end{align*}

  • Holders’ inequality: \({\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_p {\left\lVert {f} \right\rVert}_q\)

  • Riesz Representation for \(L^2\): If \(\Lambda \in (L^2) {}^{ \vee }\) then there exists a unique \(g\in L^2\) such that \(\Lambda(f) = \int fg\).

  • \({\left\lVert {f} \right\rVert}_{L^\infty(X)} \coloneqq\inf \left\{{t\geq 0 {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} \leq t \text{ almost everywhere} }\right\}\).

  • Lemma: \(m(X) < \infty \implies L^p(X) \subset L^2(X)\).

-   Write Holder's inequality as ${\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_a {\left\lVert {g} \right\rVert}_b$ where $\frac 1 a + \frac 1 b = 1$, then
    <span class="math display">
    {\left\lVert {f} \right\rVert}_p^p = {\left\lVert {{\left\lvert {f} \right\rvert}^p} \right\rVert}_1 \leq {\left\lVert {{\left\lvert {f} \right\rvert}^p} \right\rVert}_a ~{\left\lVert {1} \right\rVert}_b

-   Now take $a = \frac 2 p$ and this reduces to
    <span class="math display">
    {\left\lVert {f} \right\rVert}_p^p &\leq {\left\lVert {f} \right\rVert}_2^p ~m(X)^{\frac 1 b} \\
    \implies {\left\lVert {f} \right\rVert}_p &\leq {\left\lVert {f} \right\rVert}_2 \cdot O(m(X)) < \infty

  • Note \(X = [0, 1] \implies m(X) = 1\).

  • By Holder’s inequality with \(p=q=2\), \begin{align*} {\left\lVert {f} \right\rVert}_1 = {\left\lVert {f\cdot 1} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_2 \cdot {\left\lVert {1} \right\rVert}_2 = {\left\lVert {f} \right\rVert}_2 \cdot m(X)^{\frac 1 2} = {\left\lVert {f} \right\rVert}_2, \end{align*}

  • Thus \(L^2(X) \subseteq L^1(X)\)

  • Since they share a common dense subset (simple functions), \(L^2\) is dense in \(L^1\)

Let \(\Lambda \in L^1(X) {}^{ \vee }\) be arbitrary.

Let \(f\in L^2\subseteq L^1\) be arbitrary.

Claim: \(\Lambda\in L^1(X) {}^{ \vee }\implies \Lambda \in L^2(X) {}^{ \vee }\).

  • Suffices to show that \({\left\lVert {\Gamma} \right\rVert}_{L^2(X) {}^{ \vee }} \coloneqq\sup_{{\left\lVert {f} \right\rVert}_2 = 1} {\left\lvert {\Gamma(f)} \right\rvert} < \infty\), since bounded implies continuous.

  • By the lemma, \({\left\lVert {f} \right\rVert}_1 \leq C{\left\lVert {f} \right\rVert}_2\) for some constant \(C \approx m(X)\).

  • Note \begin{align*}{\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} \coloneqq\displaystyle\sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\Lambda(f)} \right\rvert}\end{align*}

  • Define \(\widehat{f} = {f\over {\left\lVert {f} \right\rVert}_1}\) so \({\left\lVert {\widehat{f}} \right\rVert}_1 = 1\)

  • Since \({\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }}\) is a supremum over all \(f \in L^1(X)\) with \({\left\lVert {f} \right\rVert}_1 =1\), \begin{align*} {\left\lvert {\Lambda(\widehat{f})} \right\rvert} \leq {\left\lVert {\Lambda} \right\rVert}_{(L^1(X)) {}^{ \vee }} ,\end{align*}

  • Then \begin{align*} \frac{{\left\lvert {\Lambda(f)} \right\rvert}}{{\left\lVert {f} \right\rVert}_1} &= {\left\lvert {\Lambda(\widehat{f})} \right\rvert} \leq {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} \\ \implies {\left\lvert {\Lambda(f)} \right\rvert} &\leq {\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }} \cdot {\left\lVert {f} \right\rVert}_1 \\ &\leq {\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }} \cdot C {\left\lVert {f} \right\rVert}_2 < \infty \quad\text{by assumption} ,\end{align*}

  • So \(\Lambda \in (L^2) {}^{ \vee }\).

Now apply Riesz Representation for \(L^2\): there is a \(g \in L^2\) such that \begin{align*}f\in L^2 \implies \Lambda(f) = {\left\langle {f},~{g} \right\rangle} \coloneqq\int_0^1 f(x) \overline{g(x)}\, dx.\end{align*}

  • It suffices to show \({\left\lVert {g} \right\rVert}_{L^\infty(X)} < \infty\).

  • Since we’re assuming \({\left\lVert {\Gamma} \right\rVert}_{L^1(X) {}^{ \vee }} < \infty\), it suffices to show the stated equality.

    \todo[inline]{Is this assumed..? Or did we show it..?}

  • Claim: \({\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} ={\left\lVert {g} \right\rVert}_{L^\infty(X)}\)

    • The result will follow since \(\Lambda\) was assumed to be in \(L^1(X) {}^{ \vee }\), so \({\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} < \infty\).

    • \(\leq\): \begin{align*} {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\Lambda(f)} \right\rvert} \\ &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lvert {\int_X f \overline{g}} \right\rvert} \quad\text{by (i)}\\ &= \sup_{{\left\lVert {f} \right\rVert}_1 = 1} \int_X {\left\lvert {f \overline{g}} \right\rvert} \\ &\coloneqq\sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lVert {fg} \right\rVert}_1 \\ &\leq \sup_{{\left\lVert {f} \right\rVert}_1 = 1} {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_\infty \quad\text{by Holder with } p=1,q=\infty\\ &= {\left\lVert {g} \right\rVert}_\infty ,\end{align*}

    • \(\geq\):

      • Suppose toward a contradiction that \({\left\lVert {g} \right\rVert}_\infty > {\left\lVert {\Lambda} \right\rVert}_{1 {}^{ \vee }}\).

      • Then there exists some \(E\subseteq X\) with \(m(E) > 0\) such that \begin{align*}x\in E \implies {\left\lvert {g(x)} \right\rvert} > {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }}.\end{align*}

      • Define \begin{align*} h = \frac{1}{m(E)} \frac{\overline{g}}{{\left\lvert {g} \right\rvert}} \chi_E .\end{align*}

      • Note \({\left\lVert {h} \right\rVert}_{L^1(X)} = 1\).

      • Then \begin{align*} \Lambda(h) &= \int_X hg \\ &\coloneqq\int_X \frac{1}{m(E)} \frac{g \overline g}{{\left\lvert {g} \right\rvert}} \chi_E \\ &= \frac{1}{m(E)} \int_E {\left\lvert {g} \right\rvert} \\ &\geq \frac{1}{m(E)} {\left\lVert {g} \right\rVert}_\infty m(E) \\ &= {\left\lVert {g} \right\rVert}_\infty \\ &> {\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }} ,\end{align*} a contradiction since \({\left\lVert {\Lambda} \right\rVert}_{L^1(X) {}^{ \vee }}\) is the supremum over all \(h_\alpha\) with \({\left\lVert {h_\alpha} \right\rVert}_{L^1(X)} = 1\).

Spring 2016.6 #real_analysis/qual/work

Without using the Riesz Representation Theorem, compute \begin{align*} \sup \left\{\left|\int_{0}^{1} f(x) e^{x} d x\right| {~\mathrel{\Big\vert}~}f \in L^{2}([0,1], m),~~ \|f\|_{2} \leq 1\right\} \end{align*}

Spring 2015.5 #real_analysis/qual/work

Let \(\mathcal H\) be a Hilbert space.

  • Let \(x\in \mathcal H\) and \(\left\{{u_n}\right\}_{n=1}^N\) be an orthonormal set. Prove that the best approximation to \(x\) in \(\mathcal H\) by an element in \(\mathop{\mathrm{span}}_{\mathbf{C}}\left\{{u_n}\right\}\) is given by \begin{align*} \widehat{x} \coloneqq\sum_{n=1}^N {\left\langle {x},~{u_n} \right\rangle}u_n. \end{align*}
  • Conclude that finite dimensional subspaces of \(\mathcal H\) are always closed.

Fall 2015.6 #real_analysis/qual/work

Let \(f: [0, 1] \to {\mathbf{R}}\) be continuous. Show that \begin{align*} \sup \left\{\|f g\|_{1} {~\mathrel{\Big\vert}~}g \in L^{1}[0,1],~~ \|g\|_{1} \leq 1\right\}=\|f\|_{\infty} \end{align*}

Fall 2014.6 #real_analysis/qual/work

Let \(1 \leq p,q \leq \infty\) be conjugate exponents, and show that \begin{align*} f \in L^p({\mathbf{R}}^n) \implies \|f\|_{p} = \sup _{\|g\|_{q}=1}\left|\int f(x) g(x) d x\right| \end{align*}

Banach and Hilbert Spaces

Fall 2021.5 #real_analysis/qual/work

Consider the Hilbert space \(\mathcal{H}=L^{2}([0,1])\).

  • Prove that of \(E \subset \mathcal{H}\) is closed and convex then \(E\) contains an element of smallest norm.

    Hint: Show that if \(\left\|f_{n}\right\|_{2} \rightarrow \min \left\{f \in E:\|f\|_{2}\right\}\) then \(\left\{f_{n}\right\}\) is a Cauchy sequence.

  • Construct a non-empty closed subset \(E \subset \mathcal{H}\) which does not contain an element of smallest norm.

Spring 2019.1 #real_analysis/qual/completed

Let \(C([0, 1])\) denote the space of all continuous real-valued functions on \([0, 1]\).

  • Prove that \(C([0, 1])\) is complete under the uniform norm \({\left\lVert {f} \right\rVert}_u := \displaystyle\sup_{x\in [0,1]} |f (x)|\).

  • Prove that \(C([0, 1])\) is not complete under the \(L^1{\hbox{-}}\)norm \({\left\lVert {f} \right\rVert}_1 = \displaystyle\int_0^1 |f (x)| ~dx\).

\todo[inline]{Add concepts.}

  • Let \(\left\{{f_n}\right\}\) be a Cauchy sequence in \(C(I, {\left\lVert {{-}} \right\rVert}_\infty)\), so \(\lim_n\lim_m {\left\lVert {f_m - f_n} \right\rVert}_\infty = 0\), we will show it converges to some \(f\) in this space.

  • For each fixed \(x_0 \in [0, 1]\), the sequence of real numbers \(\left\{{f_n(x_0)}\right\}\) is Cauchy in \({\mathbf{R}}\) since \begin{align*} x_0\in I \implies {\left\lvert {f_m(x_0) - f_n(x_0)} \right\rvert} \leq \sup_{x\in I} {\left\lvert {f_m(x) - f_n(x)} \right\rvert} \coloneqq{\left\lVert {f_m - f_n} \right\rVert}_\infty \overset{m>n\to\infty}\to 0, \end{align*}

  • Since \({\mathbf{R}}\) is complete, this sequence converges and we can define \(f(x) \coloneqq\lim_{k\to \infty} f_n(x)\).

  • Thus \(f_n\to f\) pointwise by construction

  • Claim: \({\left\lVert {f - f_n} \right\rVert} \overset{n\to\infty}\to 0\), so \(f_n\) converges to \(f\) in \(C([0, 1], {\left\lVert {{-}} \right\rVert}_\infty)\).

    • Proof:
      • Fix \({\varepsilon}> 0\); we will show there exists an \(N\) such that \(n\geq N \implies {\left\lVert {f_n - f} \right\rVert} < {\varepsilon}\)
      • Fix an \(x_0 \in I\). Since \(f_n \to f\) pointwise, choose \(N_1\) large enough so that \begin{align*}n\geq N_1 \implies {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} < {\varepsilon}/2.\end{align*}
      • Since \({\left\lVert {f_n - f_m} \right\rVert}_\infty \to 0\), choose and \(N_2\) large enough so that \begin{align*}n, m \geq N_2 \implies {\left\lVert {f_n - f_m} \right\rVert}_\infty < {\varepsilon}/2.\end{align*}
      • Then for \(n, m \geq \max(N_1, N_2)\), we have \begin{align*} {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &= {\left\lvert {f_n(x_0) - f(x_0) + f_m(x_0) - f_m(x_0)} \right\rvert} \\ &= {\left\lvert {f_n(x_0) - f_m(x_0) + f_m(x_0) - f(x_0)} \right\rvert} \\ &\leq {\left\lvert {f_n(x_0) - f_m(x_0)} \right\rvert} + {\left\lvert {f_m(x_0) - f(x_0)} \right\rvert} \\ &< {\left\lvert {f_n(x_0) - f_m(x_0)} \right\rvert} + {{\varepsilon}\over 2} \\ &\leq \sup_{x\in I} {\left\lvert {f_n(x) - f_m(x)} \right\rvert} + {{\varepsilon}\over 2} \\ &< {\left\lVert {f_n - f_m} \right\rVert}_\infty + {{\varepsilon}\over 2} \\ &\leq {{\varepsilon}\over 2} + {{\varepsilon}\over 2} \\ \implies {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &< {\varepsilon}\\ \implies \sup_{x\in I} {\left\lvert {f_n(x_0) - f(x_0)} \right\rvert} &\leq \sup_{x\in I} {\varepsilon}\quad\text{by order limit laws} \\ \implies {\left\lVert {f_n - f} \right\rVert} &\leq {\varepsilon}\\ .\end{align*}
  • \(f\) is the uniform limit of continuous functions and thus continuous, so \(f\in C([0, 1])\).

  • It suffices to produce a Cauchy sequence that does not converge to a continuous function.

  • Take the following sequence of functions:

    • \(f_1\) increases linearly from 0 to 1 on \([0, 1/2]\) and is 1 on \([1/2, 1]\)
    • \(f_2\) is 0 on \([0, 1/4]\) increases linearly from 0 to 1 on \([1/4, 1/2]\) and is 1 on \([1/2, 1]\)
    • \(f_3\) is 0 on \([0, 3/8]\) increases linearly from 0 to 1 on \([3/8, 1/2]\) and is 1 on \([1/2, 1]\)
    • \(f_3\) is 0 on \([0, (1/2 - 3/8)/2]\) increases linearly from 0 to 1 on \([(1/2 - 3/8)/2, 1/2]\) and is 1 on \([1/2, 1]\)

    Idea: take sequence starting points for the triangles: \(0, 0 + {1\over 4}, 0 + {1 \over 4} + {1\over 8}, \cdots\) which converges to \(1/2\) since \(\sum_{k=1}^\infty{1\over 2^k} = -{1\over 2} + \sum_{k=0}^\infty {1\over 2^k}\).

  • Then each \(f_n\) is clearly integrable, since its graph is contained in the unit square.

  • \(\left\{{f_n}\right\}\) is Cauchy: geometrically subtracting areas yields a single triangle whose area tends to 0.

  • But \(f_n\) converges to \(\chi_{[{1\over 2}, 1]}\) which is discontinuous.

\todo[inline]{show that $\int_0^1 {\left\lvert {f_n(x) - f_m(x)} \right\rvert} \,dx \to 0$ rigorously, show that no $g\in L^1([0, 1])$ can converge to this indicator function.}

Spring 2017.6 #real_analysis/qual/completed

Show that the space \(C^1([a, b])\) is a Banach space when equipped with the norm \begin{align*} \|f\|:=\sup _{x \in[a, b]}|f(x)|+\sup _{x \in[a, b]}\left|f^{\prime}(x)\right|. \end{align*}

\todo[inline]{Add concepts.}

  • Denote this norm \({\left\lVert {{-}} \right\rVert}_u\)

  • Let \(f_n\) be a Cauchy sequence in this space, so \({\left\lVert {f_n} \right\rVert}_u < \infty\) for every \(n\) and \({\left\lVert {f_j - f_k} \right\rVert}_u \overset{j, k\to\infty}\to 0\).

and define a candidate limit: for each \(x\in I\), set \begin{align*}f(x) \coloneqq\lim_{n\to\infty} f_n(x).\end{align*}

  • Note that \begin{align*} {\left\lVert {f_n} \right\rVert}_\infty &\leq {\left\lVert {f_n} \right\rVert}_u < \infty \\ {\left\lVert {f_n'} \right\rVert}_\infty &\leq {\left\lVert {f_n} \right\rVert}_u < \infty .\end{align*}

    • Thus both \(f_n, f_n'\) are Cauchy sequences in \(C^0([a, b], {\left\lVert {{-}} \right\rVert}_\infty)\), which is a Banach space, so they converge.
  • So

    • \(f_n \to f\) uniformly (by uniqueness of limits),
    • \(f_n' \to g\) uniformly for some \(g\), and
    • \(f, g\in C^0([a, b])\).
  • Claim: \(g = f'\)

    • For any fixed \(a\in I\), we have \begin{align*} f_n(x) - f_n(a) \quad &\overset{u}\to f(x) - f(a) \\ \int_a^x f'_n \quad &\overset{u}\to \int_a^x g .\end{align*}
    • By the FTC, the left-hand sides are equal.
    • By uniqueness of limits so are the right-hand sides, so \(f' = g\).
  • Claim: the limit \(f\) is an element in this space.

    • Since \(f, f'\in C^0([a, b])\), they are bounded, and so \({\left\lVert {f} \right\rVert}_u < \infty\).
  • Claim: \({\left\lVert {f_n - f} \right\rVert}_u \overset{n\to\infty}\to 0\)

  • Thus the Cauchy sequence \(\left\{{f_n}\right\}\) converges to a function \(f\) in the \(u{\hbox{-}}\)norm where \(f\) is an element of this space, making it complete.

Fall 2017.6 #real_analysis/qual/work

Let \(X\) be a complete metric space and define a norm \begin{align*} \|f\|:=\max \{|f(x)|: x \in X\}. \end{align*}

Show that \((C^0({\mathbf{R}}), {\left\lVert {{-}} \right\rVert} )\) (the space of continuous functions \(f: X\to {\mathbf{R}}\)) is complete.

\todo[inline]{Add concepts.}
\todo[inline]{Shouldn't this be a supremum? The max may not exist?}
\todo[inline]{Review and clean up.}

Let \(\left\{{f_k}\right\}\) be a Cauchy sequence, so \({\left\lVert {f_k} \right\rVert} < \infty\) for all \(k\). Then for a fixed \(x\), the sequence \(f_k(x)\) is Cauchy in \({\mathbf{R}}\) and thus converges to some \(f(x)\), so define \(f\) by \(f(x) \coloneqq\lim_{k\to\infty} f_k(x)\).

Then \({\left\lVert {f_k - f} \right\rVert} = \max_{x\in X}{\left\lvert {f_k(x) - f(x)} \right\rvert} \overset{k\to\infty}\to 0\), and thus \(f_k \to f\) uniformly and thus \(f\) is continuous. It just remains to show that \(f\) has bounded norm.

Choose \(N\) large enough so that \({\left\lVert {f - f_N} \right\rVert} < \varepsilon\), and write \({\left\lVert {f_N} \right\rVert} \coloneqq M < \infty\)

\begin{align*} {\left\lVert {f} \right\rVert} \leq {\left\lVert {f - f_N} \right\rVert} + {\left\lVert {f_N} \right\rVert} < \varepsilon + M < \infty .\end{align*}

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