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\begin{document}
\begin{enumerate}
\centerline{\sc Math 597.  Homework 12\\ Bryden Cais}
\medskip

\item Suppose that $f_j\rightarrow f$ in $L^p$ and $g_j\rightarrow g$ in $L^q$ with $1/p+1/q=1$ and $p\geq 1$.
Then we have
\begin{align*}
    \left|\int f_jg_j-fg \,\mathrm{d}\mu\right| &\leq \int |f_jg_j-f_jg+f_jg-fg|\,\mathrm{d}\mu \\
    &\leq \int |f_j||g_j-g|\,\mathrm{d}\mu+ \int |g||f_j-f|\,\mathrm{d}\mu\\
    &\leq \|f_j\|_p\|g_j-g\|_q + \|g\|_q\|f_j-f\|_p,
\end{align*}
where we have used Minkowski's inequality.  However, since $f_j\rightarrow f$ and $g_j\rightarrow g$ in $L^p,L^q$ respectively,
we have $\|f_j\|_p \leq C$ and $\|g\|\leq D$ for all sufficiently large $j$ and some constants $C,D$ (since
$\|f-f_j\|_p$ can be made arbitrarily small by making $j$ suff. large and since $\|g\|<\infty$).  Moreover, for any $\e>0$
there exists $N$ such that for all $j>N$ we have $\|g-g_j\|_q <\e/C$ and $\|f_j-f\|<\e/D$ so that for all sufficiently
large $j$ we have $$\left|\int f_jg_j\,\mathrm{d}\mu-\int fg\,\mathrm{d}\mu\right|<\epsilon$$ as required.

\item
\begin{enumerate}
    \item Let $(X,\mathcal{A},\mu)$ be a finite measure space and $f\in L^2(X,\mu)$.
    Let $f_1=\frac{1}{\sqrt{\mu(X)}}\chi_X$ (which is well defined and in $L^2$ since $X$ is finite) and observe that
    $\langle f_1,f_1\rangle=1$.  From class, we have the formula
    $$\|f-\alpha f_1\|^2=\|f\|^2-\|\langle f,f_1\rangle\|^2+\|\langle f,f_1\rangle-\alpha\|$$
    so that $\|f-\alpha f_1\|=\|f-\frac{\alpha}{\sqrt{\mu(X)}}\|:=\|f-k\|$ is minimized when
    $\alpha=\langle f,f_1\rangle$, that is, when
    $$k= \frac{1}{\mu(X)}\int_X f\,\mathrm{d}\mu.$$

    \item Let $f_j=\frac{1}{2^{-n/2}}\chi_{I_j}$ where $I_j$ is the $j^{\text{th}}$ dyadic interval of length $2^{-n}$ contained in $[0,1]$.
    It is clear that $\{f_j\}$ is orthonormal, so that we may apply the result from class as above to see that
    $$\|f-\sum_{j=1}^{2^n} \alpha_j f_j\|$$ is minimized when $\alpha_j=\langle f,f_j\rangle$.  We thus conclude that
    $$\|f-\sum_{j=1}^{2^n} c_j \chi_{I_j}\|$$ is minimized when
    $$c_j=2^n \int_{I_j} f\,\mathrm{d}\mu.$$

    \item Observe that the span of the first $2^n$ Haar functions is exactly the span of the $\chi_{I_j}$ for
    $I_j$ the dyadic intervals in $[0,1]$ of length $2^{-n}$.  Now, from class, we know that
    $$\sum_{j=1}^{2^n}\langle f,f_j\rangle f_j$$ is the {\em best} approximation to $f$ (with respect to $\|\dot\|$)
    that we can obtain using functions in the span of the first $2^n$ $f_j$.  But in part (b) we showed that
    $$\sum_{j=1}^{2^n} c_j\chi_{I_j}$$ is the best approximation to $f$ we can obtain from functions in the span of the
    $2^n$ $\chi_{I_j}$.  Since the span of the $\chi_{I_j}$ and the span of the first $2^n$ $f_j$ agree, it follows that
    the sums
    $$\sum_{j=1}^{2^n} c_j\chi_{I_j} \quad \text{and}\quad \sum_{j=1}^{2^n} \langle f,f_j\rangle f_j$$
    are equal (as functions on $L^2([0,1],\mu_{\rm Leb})$).

\end{enumerate}

\item Write
$$n\chi_{[0,1/n]}=\sum_{j\in\Z} a_n e^{ijx}.$$
Then we easily find that
\begin{align*}
a_j&=\frac{1}{2\pi i} \int_{0}^{2\pi} n\chi_{[0,1/n]}e^{-ijx}\,\mathrm{d}x\\
&=\frac{n}{2\pi}\int_0^{1/n} e^{-ijx}\,\mathrm{d}x\\
&=\frac{n}{2\pi i j}(1-e^{-ij/n}),
\end{align*}
if $j\neq 0$ and
$$a_0=\frac{n}{2\pi}\int_0^{1/n}\,\mathrm{d}x=\frac{1}{2\pi},$$
so that
$$n\chi_{[0,1/n]}(x)=\frac{1}{2\pi}\left(1+n\sum_{j\neq 0} \frac{1-e^{-ij/n}}{ij}e^{-ijx}\right).$$
Observe that for all $n$ we have $\int_{\R} n\chi_{[0,1/n]}=1$ and $$n\chi_{[0,1/n]}=\begin{cases} n & \text{if}\ x\in[0,1/n]\\ 0 & \text{otherwise}\end{cases},$$
so intuitively speaking, the limit function as $n\rightarrow \infty$ is just the dirac delta function $\delta(x)$
satisfying $\delta(x)=0$ for all $x\neq 0$ and $\int_R \delta=1$.

\item Let $A$ denote the annulus $\{z\in\C:r_1 <|z|<r_2\}$ with $0<r_1<1<r_2$, and $H$ the region
$\{z\in\C: -\log(r_2)<\Im(z)<-\log(r_1)\}$.  Then the map $z\mapsto \exp(iz)$ is a conformal map from
$H$ to $A$.  Any holomorphic function $h$ on $A$ pulls back to a holomorphic function $f=h\circ \exp$
on $H$.  Further, any such $h$ has a finite tailed laurent expansion converging uniformly to $h$ in the annulus;
this gives rise to a finite tailed laurent expansion (now in the uniformizer $w=e^{iz}$) for $f$ converging uniformly in $H$.
Observe that $H$ contains the real axis.  It follows that the fourier series of $f$ converges uniformly to $f$ on $\R$.

\item
\begin{enumerate}
    \item Suppose that $f$ has fourier series $\sum_{j\in\Z} c_je^{ijx}$ so that
    $$\frac{1}{2\pi}\int_{0}^{2\pi} f(x)e^{-ijx}\,\mathrm{d}x=c_j$$.
    Then we have
    \begin{align*}
    \frac{1}{2\pi}\int_{0}^{2\pi} f_{\alpha}(x)e^{-ijx}\,\mathrm{d}x&=\frac{1}{2\pi}\int_{0}^{2\pi} f(x+\alpha)e^{-ijx}\,\mathrm{d}x\\
    &=\frac{1}{2\pi}\int_{\alpha}^{2\pi+\alpha} f(x)e^{-ij(x-\alpha)}\,\mathrm{d}x \\
    &=\frac{1}{2\pi}\left(\int_{2\pi}^{2\pi+\alpha} f(x)e^{-ij(x-\alpha)}\,\mathrm{d}x+\int_{\alpha}^{2\pi} f(x)e^{-ij(x-\alpha)}\,\mathrm{d}x\right)\\
    &=\frac{1}{2\pi}\left(\int_{0}^{\alpha} f(x)e^{-ij(x-\alpha)}\,\mathrm{d}x+\int_{\alpha}^{2\pi} f(x)e^{-ij(x-\alpha)}\,\mathrm{d}x\right)\\
    \intertext{since $f$ has period $2\pi$,}
    &=\frac{e^{ij\alpha}}{2\pi}\int_{0}^{2\pi} f(x)e^{-ijx}\,\mathrm{d}x\\
    &=c_je^{ij\alpha},
    \end{align*}
    from which it follows that
    $$f_{\alpha}(x)=\sum_{j\in\Z} c_je^{ij\alpha} e^{ijx}.$$

    \item Observe that $e^{ijx}\in L^2([0,2\pi))$ for all $j\in \Z$ and that these functions are linearly independent.
    Now if $\alpha=\pi p/q\in \pi\Q$ for $p,q\in \Z$ then $e^{2qij x}=e^{2qij(x+\alpha)}$ as functions on $[0,2\pi)$.  It follows that
    if $\alpha$ is a nonzero rational multiple of $\pi$ then the dimension of $\{f\in L^2:f=f_{\alpha}\}$ is infinite.
    On the other hand, suppose that $\alpha/\pi$ is not rational.  Then $e^{ijx}\neq e^{ij(x+\alpha)}$ for any $j\in \Z$
    (for this would require $e^{ij\alpha}=1,$ or equivalently, $\alpha/\pi\in\Q$).  Since $e^{ijx}$ for $j\in \Z$ are dense
    in $L^2([0,2\pi))$, we see that the dimension of $\{f\in L^2: f=f_{\alpha}\}$ is 0.
    Finally, if $\alpha=0$ it is clear that the above space is infinite dimensional.
\end{enumerate}



\end{enumerate}
\end{document}
