Pothos

Theorem. For $p\in [1, \infty)$, $L^p$ is complete.

Proof. Pick a Cauchy sequence $f_n$. We can pick a subsequence $f_{n_k}$ such that $\|f_{n_{k+1}}- f_{n_k}\| \leq 2^{-k}$. Set $S_K(f) = f_{n_1} + \sum_{i=1}^{K} (f_{n_{i+1}} – f_{n_i})$ and also $S_K(g) = |f_{n_1}| + \sum_{i=1}^{K} |f_{n_{i+1}}- f_{n_i}|$. By the triangle inequality on the $p$-norm, $\|S_K(g)\|_p \leq \|f_{n_1}\|_p + \sum_{i=1}^K \|f_{n_{i+1}}- f_{n_i}\|_p \leq \|f_{n_1}\|_p + 1$, so by Monotone Convergence, $S_\infty(g) \in L^p$, hence $S_\infty(f)$ is also in $L^p$, hence it converges almost everywhere to some $f\in L^p$. But clearly $S_K(f) = f_{n_{K+1}}$, so $f_{n_k} \to f$ almost everywhere. You can then show using DCT that $f_{n_k} \to f$ in the $p$-norm. By the Cauchy property of $f_n$, you then also show that $f_n \to f$ in $L^p$. $\blacksquare$

Theorem. Suppose $p, q \in [1, \infty]$ are conjugates, and $fg$ is integrable for each $g\in L^q$. Then $f\in L^p$.

Proof. We can find a sequence of simple functions $s_k$ which converges to $f$ in $L^1$ with $|s_k| \leq |f|$. We may choose these such that $|s_k(x)|$ is increasing for each $x\in X$. Each $s_k$ trivially belongs to $L^p$, by simplicity. Now, let $S_k(h) = \int s_kh$ be a functional on $L^q$. Linearity is obvious. Boundedness follows from Hölder’s Inequality, and in fact the extremal form thereof implies that $\|S_k\|_* = \|s_k\|_p$.

Pick any $h\in L^q$. Then DCT and our assumption that $fh\in L^1$ implies $\lim_k \int s_kh = \int \lim_k s_kh = \int fh \in \Bbb C$, hence $\sup_k \{ |S_k(h)| \} < \infty$. As $h$ was arbitrary, Uniform Boundedness implies that $\sup_k \{ \|S_k\|_*\} = \sup_k \{ \|s_k\|_p\}< \infty$. However, $\lim_k \int |s_k|^p = \int |f|^p$ by Monotone Convergence, so $f\in L^p$. $\blacksquare$

Theorem. Suppose $p\in (1, \infty)$ and let $D = (L^p)^*$, the dual of $L^p$. Then $D$ is isomorphic to $L^q$, where $1/p + 1/q = 1$. (We’re assuming an ambient $\sigma$-finite measure space $(X, \Sigma, \mu)$ here.)

Proof. D is of course already a Banach space with norm

$$\|T\|_D = \inf \{ M : |Tf| \leq M \|f\|_p \text{ for all } f\in L^p \}.$$

We want to map the bounded linear functional $T$ on $L^p$ to some function $g_T \in L^q$ such that $\|T\|_D = \|g_T\|_q.$ 

Let’s first show that $T\in D$ implies a $g$ such that $Tf = \int fg$. When $f$ is the characteristic function of some measurable set $A$, we can define a complex set function by $\nu(A) = Tf$. By linearity, $\nu$ is already finitely additive. On the other hand, take a countable disjoint collection $A_i$ and define $B_i = \bigcup_{k=i}^\infty A_i$; assume $1_{B_0}\in L^p$. Finite additivity implies that for each $k\in\Bbb N$,

$$\nu(B_0) = \nu(A_0) + \ldots + \nu(A_k) + \nu(B_{k+1})\tag{1}.$$

But $\nu(B_k) = T1_{B_k}$. Boundedness of $T$ implies

$$\begin{eqnarray}
|T1_{B_k}| &\leq& \|T\|_D \|1_{B_k}\|_p \\
&=& \|T\|_D \left( \int_{B_k} d\mu \right)^{1/p}.
\end{eqnarray}$$

Now, $B_k \to \emptyset$, and since $B_0 \supseteq B_k$ is of finite measure (as $1_{B_0} \in L^p$), $\int_{B_k} d\mu \to 0$. Therefore $\nu(B_k) \to 0$, so taking limits of $(1)$ establishes countable additivity of $\nu$. Finally, for any $\mu$-null set $A$, $|\nu(A)| = |T1_A| \leq \|T\|_D \left( \int_A d\mu \right)^{1/p} = 0$, hence $\nu \ll \mu$. So Radon-Nikodym implies that $T1_A = \nu(A) = \int_A g\ d\mu$. It is not difficult to see that this, together with continuity of $T$, implies $Tf = \int gf\ d\mu$ for $f\in L^p$. Our previous theorem then implies that $g\in L^q$.

Hölder’s Inequality thus implies that $|Tf|\leq \int |fg|\leq \|f\|_p \|g\|_q$, so that $\|T\|_D \leq \|g\|_q$. Hölder’s Inequality is also sharp, so we can select $f$ to make this an equality.

Lemma 1. Suppose $f$ is continuous and of moderate decrease, i.e., $|f(x)| <= {A \over (1 + x^2)}$. Then its periodization $g(x) = \sum_{k\in\Bbb Z} f(x + k)$ is continuous and periodic, and its partial sums converge uniformly to $g$.

Proof. Let $S_n(x) = \sum_{|k| \leq n} f(x + k)$.  Given $n > m$, we observe the Cauchy sum of $|S_n – S_m|$ is bounded by $\sum_{|k| > n} |f(x + k)|$. But for $x \in [0, 1)$,

$$
\begin{eqnarray}
\sum_{|k| > n} |f(x + k)| &\leq& A\sum_{|k| > n} \frac{1}{1 + (x + k)^2} \\
&=& A\sum_{k > n} \frac{1}{1 + (x + k)^2} + A\sum_{k < -n} \frac{1}{1 + (x + k)^2} \\
&\leq& A\sum_{k > n} \frac{1}{1 + k^2} + A\sum_{k \leq -n} \frac{1}{1 + k^2} \\
&\leq& A\sum_{k > n} \frac{1}{k^2} + A\sum_{k \leq -n} \frac{1}{k^2} \\
&\leq& 2AT_n
\end{eqnarray}
$$

where $T_n$ is the tail sum of $\zeta(2)$. Hence $S_n$ is uniformly Cauchy on $[0, 1)$, so it converges uniformly to $g$ there, so $g$ is continuous on that interval. But $g$ is obviously periodic from its form, so the uniform convergence and continuity apply everywhere. (Possibly more details needed here.) $\blacksquare$

Lemma 2. If $|f(x+iy)| \leq \frac{A}{1 + x^2}$ for some constant $A$ on some strip $-a \leq y \leq a$, and f is holomorphic on this strip, then $|\hat f|$ decreases exponentially.

Proof. We integrate along a box from $-R \to R \to R- ai \to -R- ai \to -R$. The sides go to $0$ with $R$ and $f$ is holomorphic on the interior, so the top and bottom in the limit are equal and we have

$$
\begin{eqnarray}
\int_{-\infty}^\infty f(t)e^{-2\pi is t} dt &=& \int_{-\infty-ai}^{\infty-ai} f(t)e^{-2\pi is t} dt \\
&=& \int_{-\infty}^{\infty} f(t- ai)e^{-2\pi is (t – ai)} dt \\
&=& e^{-2\pi a s} \int_{-\infty}^{\infty} f(t- ai)e^{-2\pi is} dt
\end{eqnarray}
$$

and this is of course bounded in magnitude by $Ce^{-2\pi a s}$. $\blacksquare$

Theorem.Suppose $f$ is continuous and of moderate decrease with Fourier transform $\hat f$ and periodization $g$. Then the $n$th Fourier coefficient of $g$ is given by $\hat f(n)$.

Proof.

$$
\begin{eqnarray}
\frac{1}{2\pi} \int_{-\pi}^\pi g(t)e^{-ikt} dt &=& \frac{1}{2\pi} \int_0^{2\pi} g(t)e^{-ikt} dt \\
&=& \int_{0}^1 g(t)e^{-2\pi ikt} dt \\
&=& \int_{0}^1 \sum_{n \in \Bbb Z} f(t + n) e^{-2\pi ikt} dt \\
&=& \sum_{n \in \Bbb Z} \int_0^1 f(t + n) e^{-2\pi ikt} dt \\
&=& \sum_{n \in \Bbb Z} \int_n^{n+1} f(t) e^{-2\pi ikt} dt \\
&=&  \int_{-\infty}^\infty f(t) e^{-2\pi ikt} dt \\
&=& \hat f(n).
\end{eqnarray}
$$

The interchange of integral and sum is justified by the uniform convergence, as established above. $\blacksquare$

It follows that $\sum_{n\in \Bbb Z}f(t + n) \sim \sum_{n\in \Bbb Z} \hat f(n) e^{2\pi int}$. If $f$ is decreases fast enough  – e.g., if it’s Schwartz, or satisfies the conditions of Lemma 2. above – then this is actually an equality, and we can plug in $t=0$ to yield $\sum_{n\in \Bbb Z}f(n) \sim \sum_{n\in \Bbb Z} \hat f(n)$.

Theorem (Minkowski’s Inequality). If $f, g \in L^p$ for $1 \leq p < \infty$, $\|f + g\|_p \leq \|f\|_p + \|g\|_p$.

Proof. Start by noticing that $|f + g|^p \leq 2^{p-1}(|f|^p + |g|^q)$ by Jensen’s inequality, so that $\|f + g\|_p$ exists and is finite.

Let $1/p + 1/q = 1$, which is equivalent to $p + q = pq$. Notice that

$$
\begin{eqnarray}
\|(f+g)^{p-1}\|_q &=& \left( \int \left(|f + g|^{p-1}\right)^q d\mu \right)^{1/q} \\
&=& \left( \int |f + g|^{pq-q} d\mu \right)^{1/q} \\
&=& \left( \int |f + g|^p d\mu \right)^{1/q} \\
&=& \| f+g \|_p^{p/q}
\end{eqnarray}
$$

Now, by Hölder’s inequality, we see that $\| f(f+g)^{p-1} \|_1 \leq \|f\|_p\cdot \|(f+g)^{p-1}\|_q$, and similarly $\| g(f+g)^{p-1} \|_1 \leq \|g\|_p\cdot \|(f+g)^{p-1}\|_q$. Ergo

$$
\begin{eqnarray}
(\|f\|_p + \|g\|_p)\cdot \|(f+g)\|_p^{p/q} &=& (\|f\|_p + \|g\|_p) \cdot \|(f+g)^{p-1}\|_q \\
&\geq& \| f(f+g)^{p-1} \|_1 + \| g(f+g)^{p-1} \|_1 \\
&=& \int (|f| + |g|)|f + g|^{p-1} d\mu \\
&\geq& \int |f + g| \cdot |f + g|^{p-1} d\mu \\
&=& \int |f + g|^p d\mu \\
&=& \| f+g \|_p^p.
\end{eqnarray}
$$

Dividing the leftmost expression through by $\|(f+g)\|_p^{p/q}$ gives us one side of Minkowski’s inequality. Dividing through the rightmost gives us $\| f+g \|_p^{p – p/q} = \| f+g \|_p$, which is the other. $\blacksquare$

Theorem. For $p\in (1, \infty)$, $L^p$ is complete.

Proof. Pick a Cauchy sequence $f_n$. We can pick a subsequence $f_{n_k}$ such that $\|f_{n_{k+1}}- f_{n_k}\| \leq 2^{-k}$. Set $S_K(f) = f_{n_1} + \sum_{i=1}^{K} (f_{n_{i+1}} – f_{n_i})$ and also $S_K(g) = |f_{n_1}| + \sum_{i=1}^{K} |f_{n_{i+1}}- f_{n_i}|$. By the triangle inequality on the $p$-norm, $\|S_K(g)\|_p \leq \|f_{n_1}\|_p + \sum_{i=1}^K \|f_{n_{i+1}}- f_{n_i}\|_p \leq \|f_{n_1}\|_p + 1$, so by Monotone Convergence, $S_\infty(g) \in L^p$, hence $S_\infty(f)$ is also in $L^p$, hence it converges almost everywhere to some $f\in L^p$. But clearly $S_K(f) = f_{n_{K+1}}$, so $f_{n_k} \to f$ almost everywhere. You can then show using DCT that $f_{n_k} \to f$ in the $p$-norm. By the Cauchy property of $f_n$, you then also show that $f_n \to f$ in $L^p$.

Theorem. Let $B$ be a Banach space. Then its dual $C$ is also a Banach space.

Proof. Let $x \in B$ and $T, S \in C$; we define $(T + S)(x) = Tx + Sx, (\alpha T)x = \alpha(Tx)$. We define $\| T \|_C = \inf\ \{ M :|Tx| \leq M \|x\|_B \text{ for all } x \in B \}$. (We used $|Tx|$ because this is a scalar.) This of course implies $|Tx| \leq \|T\|_C \|x\|_B$. This norm obeys the triangle inequality because for all $x\in B$,

$$
\begin{eqnarray}
|(T + S)x| &=& |Tx + Sx| \\
&\leq& |Tx| + | Sx| \\
&\leq& \|T\|_C \|x\|_B + \|S\|_C \|x\|_B \\
&=& (\|T\|_C + \|S\|_C) \|x\|_B
\end{eqnarray}
$$

and $\|T+S\|_C$ is the infimum of all constants with this property.

Suppose $T_n\in C$ is a Cauchy sequence of bounded linear functionals on $B$. For all $x\in B$, our definitions imply that $|T_n x- T_m x| \leq \|T_n – T_m\|_C \| x\|_B$, so since the first factor can be made arbitrarily small for sufficiently large indices, it follows that $T_n x$ is a Cauchy sequence on $B$’s scalar field, hence it converges to some scalar we’ll call $T x$. Thus we’ve shown $T_n$ converges to a functional $T$. Clearly $T$ is linear.

We need to show that $T$ is bounded. Note that by the reverse triangle inequality, $\|T_n – T_m\|_C \geq |\|T_n\|_C- \|T_m\|_C|$. Since the left side can be made arbitrarily small for all large indices by our original assumption, so can the right side, hence $\|T_n\|_C$ forms a Cauchy sequence, so it converges to some constant. It follows, therefore, that $\|T_n\|_C$ are all uniformly bounded by some $M$. So $|Tx| \leq \|T – T_n\|_C \|x\|_B + \|T_n\|_C \|x\|_B$. Convergence of $T_n$ to $T$ in the $C$-norm implies the left term can be made arbitrarily small regardless of $x$ by appropriate choice of $n$, and the right term is bounded by $M\|x\|_B$. It follows that $\|T\|_C \leq M$ and so $T$ is bounded.

Theorem (Argument Principle). Suppose $f$ is meromorphic on an open region $R$ enclosed by a contour $C$. Then $\displaystyle \frac{1}{2\pi i} \int_C \frac{f'(z)}{f(z)} dz = N – M$, where $N$ is the number of zeroes (with multiplicity) enclosed by $C$ and $M$ is the number of poles (with multiplicity) enclosed by $C$.

Proof. At any point $z$ where $f(z)$ is holomorphic and nonzero, then the integrand is also holomorphic. So the only possible poles of the integrand occur at the zeros and the poles of $f$.

Let’s look at the residues of the zeroes first. Around any zero $z_0 \in R$ of multiplicity $m$, we can write $f(z) =(z – z_0)^mg(z)$, where $g$ is holomorphic and nonzero on some closed disk $E$ centered on $z_0$. Then $f'(z) = m(z – z_0)^{m-1}g(z) + (z – z_0)^mg'(z)$ and so $\frac{f'(z)}{f(z)} = \frac{m}{z – z_0} + \frac{g'(z)}{g(z)}$. The second term of this sum is holomorphic on $E$ and so the requisite integral around $E$ just equals $m$.

Next, let’s suppose $z_0$ is a pole of $f$ of order $m$. We can then write $f(z) = (z – z_0)^{-m}g(z)$ on some closed disk $E$ as before. Differentiating, we now get

$$
f'(z) = -m(z – z_0)^{-m-1}g(z) + (z – z_0)^{-m}g'(z)
$$

and so $\frac{f'(z)}{f(z)} = \frac{-m}{z – z_0} + \frac{g'(z)}{g(z)}$. This makes the integral around $E$ equal $-m$ (as $g$ is again holomorphic and nonzero).

The Residue Theorem then gives us the result. $\blacksquare$

Theorem (Rouché). Suppose $f$ is holomorphic on some compact region $R$ and $|g| < |f|$. Then $f + g$ has the same number of zeros as $f$ inside $R$.

Proof. Let $h_a(z) = \frac{f’ + g’}{f + ag}$ for $a \in [0, 1]$. It can be shown through simple algebra that $h_x(z) – h_y(z) = (y- x)g\frac{f’+ g’}{(f + xg)(f + yg)}$. Our assumptions about the behavior of $f$ and $g$ on the boundary ensure that the latter fraction can be bounded in magnitude by some constant $M$ uniformly in $z$. Hence we have $|\int_{\partial R} h_x – h_y dz| \leq |x – y|M l(\partial R)$ which implies continuity of $H(x) = \int_{\partial R} h_x dz$ as a function of $x$. But $H(x)$ is integer-valued, so it’s constant, so it equals $H(0)$, which is the number of zeros of $f$ in $R$.

Statement. $F[e^{-\pi x^2}] = e^{-\pi x^2}$

Proof.

$$
\begin{eqnarray}
F[e^{-\pi x^2}](s) &=& \int_{\Bbb R} e^{-\pi x^2} e^{-2\pi i sx} dx \\
&=& \int_{\Bbb R} e^{-\pi x^2 -2\pi i sx} dx \\
&=& e^{\pi (is)^2} \int_{\Bbb R} e^{-\pi x^2 -2\pi i sx – \pi(i s)^2} dx \\
&=& e^{-\pi s^2} \int_{\Bbb R} e^{-\pi(x+i s)^2} dx.
\end{eqnarray}
$$

We can approach this integral via contour integration. Draw the box from $-R \to R \to R + is \to -R + is \to -R$. On the sides, we have $\int_{R}^{R+is} e^{-\pi (x+is)^2} dx$, which gets small when $R$ increases. Thus the top side and the bottom sides cancel out in the limit since the integral over the whole contour vanishes, so $\int_{\Bbb R} e^{-\pi(x+i s)^2} = \int_{\Bbb R} e^{-\pi x^2} = 1$.

Given a periodic $f$, we know we can approximate it in $L^2$ with trigonometric polynomials. In particular, $S_n$ converges to $f$ in $L^2$. Now, $\langle \cdot, c \rangle$ is a linear operator, and by the Cauchy-Schwarz inequality, it’s bounded, hence it’s continuous. Thus, by $S_n \to f$, we have $\langle S_n, f \rangle \to \langle f, f \rangle$.

Suppose $H$ is a Hilbert space with basis $\{e_n\}_{n\in \Bbb N}$. Note that

$$\langle f-\langle f, e_k \rangle e_k, \langle f, e_k \rangle e_k \rangle = |\langle f, e_k \rangle |^2 – |\langle f, e_k \rangle|^2 = 0.$$

Therefore $\langle f – S_n, S_n \rangle = 0$. But $\langle f – S_n, S_n \rangle = \langle f, S_n \rangle – \langle S_n, S_n \rangle \to \langle f, f \rangle – \sum_{k\in \Bbb Z} |a_k|^2$. Therefore $\sum_{k \in \Bbb Z} |a_k|^2 = \| f \|^2$.

Alternately, orthogonality implies by the Pythagorean Theorem that $\| f \|^2 = \| f- S_n\|^2 + \|S_n\|^2$. Taking the limit in $n$ gets us the same result.

A sequence of kernels $K_n(x)$ on $[-\pi, \pi]$ is called *good* if

1. $\frac{1}{2\pi} \int_{-\pi}^\pi K_n(x) \, dx = 1$,
2. $\frac{1}{2\pi} \int_{-\pi}^\pi |K_n(x)| \, dx$ is bounded, and
3. $\frac{1}{2\pi} \int_{|x| \geq c} |K_n(x)| \, dx \to 0$ as $n \to \infty$ for each $c> 0$.

Theorem. Suppose $|f| \leq B$ on $[-\pi, \pi]$, $K_n$ are good kernels and $f$ is continuous at $x_0$, then $(K_n\star f)(x_0) \to f(x)$, where $\star$ represents convolution.

Proof. We can choose some bound $\int_{-\pi}^\pi |K_n(y) |\, dy \leq C$ uniform in all $n$ by the second property.  Pick $\epsilon > 0$. By continuity, there is some $\delta > 0$ such that $|y \, – x| < \delta$ implies $|f(y \, – x) – f(x)| < \epsilon$.  We have

$$
\begin{eqnarray}
2\pi \left| (K_n \star f)(x_0) – f(x_0) \right| &=& \left| \int_{-\pi}^\pi K_n(y)f(x_0 – y) \, dy \, – \int_{-\pi}^{\pi} K_n(y)f(x_0) \, dy \right| \\
&=& \left| \int_{-\pi}^\pi K_n(y)(f(x_0 – y) – f(y)) \, dy \right| \\
&\leq& \int_{-\pi}^\pi |K_n(y)| |f(x_0 – y) – f(y)| \, dy \\
&\leq& \int_{y < \delta} |K_n(y)| |f(x_0 – y) – f(y)| \, dy +
\int_{y \geq \delta} |K_n(y)| |f(x_0 – y) – f(y)| \, dy \\
&\leq& \epsilon \int_{y < \delta} |K_n(x)| \, dy + 2B \int_{y \geq \delta} |K_n(y)| \, dy \\
&\leq& \epsilon C + 2B \int_{y \geq \delta} |K_n(y)| \, dy.
\end{eqnarray}
$$

By the third property, the second term converges to zero. So the original quantity can be made arbitarily small. $\blacksquare$

Theorem. If $f$ is continuous on $[-\pi, \pi]$, then this convergence is uniform.

Proof. If $f$ is continuous on a closed interval, it’s uniformly continuous there, so the $\delta$ used in the above proof does not depend on $x_0$. $\blacksquare$

Pointwise Convergence

The Dirichlet kernel $D_n$ can be convolved with $f$ to yield the partial sums $S_n$ of the Fourier series of $f$. While it fails to obey property 2. above, we can still use it to show pointwise convergence of the Fourier series at any point where $f$ is differentiable:

$$
\begin{eqnarray}
2\pi |(D_n \star f)(x) – f(x)|
&=& \left| \int_{-\pi}^{\pi} D_n(t)f(x-t) dt -\int_{-\pi}^{\pi} D_n(t)f(x) dt \right| \\
&=& \left| \int_{-\pi}^{\pi} D_n(t)(f(x-t) – f(x)) dt \right| \\
&=& \left| \int_{-\pi}^{\pi} \frac{\sin((n+1/2)t)}{\sin(t/2)} (f(x-t) – f(x)) dt \right| \\
&=& \left| \int_{-\pi}^{\pi} \frac{f(x-t) – f(x)}{\sin(t/2)} \sin((n+1/2)t) dt \right|.
\end{eqnarray}
$$

By $f$’s differentiability at $x$ and l’Hopital’s rule, the left factor of the integrand is continuous at $t=0$, hence integrable on $[-\pi, \pi]$. Thus the Riemann-Lebesgue lemma implies the last integral tends to zero as $n \to \infty$.

In fact, this still holds if $f$ is merely Lipschitz in some neighborhood of $x$. For then if $h(t) = \frac{f(x-t)-f(x)}{\sin(t/2)}$, we have $\limsup_{t\to 0} |h(t)| \leq \limsup_{t\to 0}|Kt/\sin(t/2)| = 2K$ and so $h$ is integrable, hence Riemann-Lebesgue still applies.

The Fejer kernel $F_n(t) = \frac{1}{n}\sum_{k=0}^{n-1} D_n(t)$ is a good kernel. Hence $F_n \star f$ converges uniformly to $f$ on $[-\pi, \pi]$; but this convolution can be written as a linear combination of the form $a_n e^{inx} \int_{-\pi}^{\pi} e^{-int}f(t) dt$, it follows that all integrable functions can be uniformly approximated by trigonometric polynomials.

Theorem. Suppose $\lim_{|z|\to\infty} |f(z)| = 0$ and $a > 0$. Then $\lim_{R\to \infty} \int_{C_R} e^{iaz} f(z) \, dz = 0$ where $C_R$ is the semicircle in the upper half-plane with radius $R$ centered on $0$.

Proof. Our assumption implies that $M_R \equiv \max_{|z| = R} |f(z)|$ obeys $\lim_{R\to\infty} M_R = 0$.

Fix R and parametrize $z$ by $Re^{it} = R\cos(t) + iR\sin(t)$ for $t \in [0, \pi]$. Then we have

$$
\begin{eqnarray}
I_R &=& \int_{C_R} e^{iaz} f(z) \, dz \\
&=& \int_0^\pi e^{ia(R\cos(t) + iR\sin(t))} f(z) iRe^{it} \, dt \\
&=& iR \int_0^\pi f(z) e^{-aR\sin(t)} e^{i(t + aR\cos(t))} \, dt
\end{eqnarray}
$$

so that $|I_R| \leq RM_R \int_0^\pi e^{-aR\sin(t)} \, dt = 2RM_R \int_0^\frac{\pi}{2} e^{-aR\sin(t)} \, dt$. But note that on $[0, \frac{\pi}{2}]$ we have $\sin(t) \geq \frac{2}{\pi}t$, so the latter integral is bounded by

$$
\int_0^\frac{\pi}{2} e^{-\frac{2aR}{\pi}t} \, dt = \frac{\pi}{2aR}(1 – e^{-aR}) \leq \frac{\pi}{2aR}.
$$

Hence $|I_R| \leq 2RM_R \cdot \frac{\pi}{2aR} \to 0$. $\blacksquare$

You can also do this lemma with integrals with a negative exponent by taking the clockwise semicircle in the lower half-plane.

Let $\hat f(n) = \frac{1}{2\pi} \int_{-\pi}^\pi f(t)e^{-int} \, \text{d}t$ be the Fourier coefficients of $f$.

Theorem. $\hat f(n) = 0$ implies $f(t) = 0$ when $f$ is continuous at $t$.

Proof. By linearity, this is equivalent to showing that $\hat f(n) = 0$ for all $n$ implies $f = 0$. Assume the contrary. Then $\int_{-\pi}^{\pi} q(t)f(t) \, \text{d}t$ should always be 0, where $q$ is any trigonometric polynomial, but $f(0) \neq 0$.

The overall idea is to isolate the behavior of $f$ at $0$ via a trigonometric polynomial.

$f$ is continuous at $0$, so we can choose a $\delta < \frac{\pi}{2}$ such that $f(t) \geq f(0)/2$ when $|t| < \delta$. Pick $\epsilon > 0$ small enough that $p(t) = \epsilon + \cos(t) < 1 – \epsilon/2$ when $|t| > \delta$. Finally, choose $\eta < \delta$ with $p(t) > 1 + \epsilon/2$ when $|t| < \eta$ and set $p_k(t) = p(t)^k$.

First, $\int_{|t| \leq \eta} p_k(t)f(t)\, \text{d}t > (1 + \epsilon/2)^k f(0)/2 \to \infty$.

Second, $\int_{\eta < |t| \leq \delta} p_k(t)f(t)\, \text{d}t \geq (1 – \epsilon/2)^k f(0)/2 \geq 0$.

Third, $\left| \int_{\delta < |t| \leq \pi} p_k(t)f(t)\, \text{d}t \right| \leq (1 – \epsilon/2)^kM$ where $M$ is some bound on $f$. This quantity shrinks to 0.

Combining all three, we can see that $\int_{-\pi}^\pi p_k(t)f(t)\, \text{d}t \to \infty$ as $n$ gets large. By assumption, however, it should be zero. Contradiction. $\blacksquare$

Definition. Let $S_N(t) = \sum_{|n| \leq N}\hat f(n)e^{int}$ be the partial sums of the Fourier series of $f$.

Theorem. If $f$ is continuous and $\sum |\hat f(n)| < \infty$, then $S_N \to f$ uniformly.

Proof. As in the Weierstrass M-test to show uniform convergence (remember that $|e^{int}| = 1$) to some function $g$. Also note that the uniform limit of continuous functions is continuous, hence $g$ is continuous. Also, limiting sums commute with integrals under uniform convergence, implying

$$
\begin{eqnarray}
\hat g(n) &=& \frac{1}{2\pi} \int_{-\pi}^\pi g(t)e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \int_{-\pi}^\pi \lim_N S_N(t)e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \lim_N \int_{-\pi}^\pi S_N(t)e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \lim_N \int_{-\pi}^\pi \sum_{|m| \leq N} \hat f(m)e^{imt}e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \lim_N \sum_{|m| \leq N}\hat f(m) \int_{-\pi}^\pi e^{i(m-n)t}\, \text{d}t \\
&=& \hat f(n).
\end{eqnarray}
$$

Ergo $g = f$, since their Fourier coefficients are the same and both are continuous everywhere.

Theorem. If $f \in C^1(\Bbb T)$, then $S_n \to f$ uniformly.

Proof. $\frac{1}{2\pi} \int_{-\pi}^\pi f(t)e^{-int} dt = \frac{1}{in}\frac{1}{2\pi}\int_{-\pi}^\pi f'(t)e^{-int} dt$ through integration by parts (this is where continuity of $f’$ is needed), so that $\hat{f}(n) = \frac{1}{in} \widehat{f’}(n)$. Furthermore, $\sum |\widehat{f’}(n)|^2 =\frac{1}{2\pi} \int_{-\pi}^\pi |f'(t)|^2 dt < \infty$ by Parseval’s identity. Therefore $\sum |\hat f(n)| \leq \left( \sum \frac{1}{n^2} \right) \left( | \widehat{f’}(n)|^2 \right) < \infty$ by Cauchy-Schwarz. $\blacksquare$

I think $|c_n| \leq 1/n^{k+1+\epsilon}$ for $\epsilon > 0$ implies $f \in C_k([-\pi, \pi])$.

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