# 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: $$m(X) < \infty \implies L^p(X) \subset L^2(X)$$.

{=tex}
\hfill

-   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">
\begin{align*}
{\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
.\end{align*}
<span>{=html}

-   Now take $a = \frac 2 p$ and this reduces to

<span class="math display">
\begin{align*}
{\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
.\end{align*}
<span>{=html}

• 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.}

\hfill


• 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.}

\hfill


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