# Tuesday October 29

## Approximations of the Identity

**Theorem:**
Let $\phi \in L^1$ and $\int \phi = 1$.

Then

- If $f$ is bounded and uniformly continuous, then $f \ast \phi_t \mapsvia{u} f$ uniformly where 
$$
\phi_t(x) \definedas \frac 1 {t^n} \phi(\frac x t)
.$$

- If $f\in L^1$, then $f \ast \phi_t \mapsvia{L^1} f$ in $L_1$.


**Applications:**

## Theorem 1: Smooth Compactly Supported Functions are Dense in $L^1$

**Theorem:**
$C_c^\infty \injects L^1$ is dense,

That is, $\forall \varepsilon > 0$ and for all $f\in L^1$, there exists a $g\in C_c^\infty$ such that $\norm{f - g}_1 < \varepsilon$.

*Proof:* 
Since $C_c^0$ is dense in $L^1$, it suffices to show the following:
$$
\forall \varepsilon > 0 ~\&~ h \in C_c^1, \quad \exists g\in C_c^\infty \suchthat \norm{h - g}_1 < \varepsilon.
$$

Let $\phi \in C_c^\infty$ be arbitrary where $\int \phi = 1$ *(which exist!)*.

Then $\norm{h\ast \phi_t - h}_1 < \varepsilon$ for $t$ small enough.
It remains to show that $f\definedas h\ast \phi_t \in C_c^\infty$.

$f$ is smooth because of theorem 3 regarding convolution, applied infinitely many times.

$f$ is also compactly supported: since $h, \phi_t$ are compactly supported, so there is some large $N$ such that $\abs x > N \implies h(x) = \phi_t(x) = 0$.

Then if $\abs x > 2N$, we can note that 
$$
\abs x \leq \abs{x+y} + \abs{y},
$$
so either $\abs{x-y}\geq 2N$ or $\abs y \geq N$.
  
But then 
$$
f(x) \definedas h \ast \phi_t(x) = \int h(x-y)\phi_t(y)~dy = 0
,$$ 
where by the previous statement, at least one term in the integrand is zero and thus the integral is zero and $f \definedas h \ast \phi_t$ compactly supported.

$\qed$

## Theorem 2: Weierstrass Approximation:

**Theorem:**
A function can be *uniformly* approximated by a polynomial on any closed interval, i.e.
$$
\forall\varepsilon > 0,~ f\in C([a, b]),\quad \exists \text{ a polynomial } P \suchthat \abs{f(x) - P(x)} < \varepsilon \quad \forall x\in [a, b].
$$

*Proof:*
Let $g$ be a continuous function on $[-M, M] \supseteq [a, b]$ such that $\restrictionof{g}{[a, b]} = f$.

Let $\phi(x) = e^{-\pi x^2}$ be the standard Gaussian, then $g \ast \phi_t \uniformlyconverges g$ on $[-M, M ]$, and thus $g\ast \phi_t \uniformlyconverges f$ on $[a, b]$.

> The problem is that this is not a polynomial.

We can let $\varepsilon > 0$, then there is a $t$ such that 
$$
\abs{g\ast \phi_t(x) - g(x)} < \varepsilon \quad \forall x\in[-M, M]
.$$

Note that $\phi_t(x) = \frac 1 t e^{-\pi x^2 / t^2}$, and Maclaurin expand to obtain 
$$
P(t) \definedas \frac 1 t \sum_{n=0}^\infty \frac{(-1)^n \pi^n x^{2n}}{t^{2n} n!}
.$$

> Note that the Maclaurin series will converge uniformly on compact sets!

By uniform convergence of $P$, we can truncate it to bound the difference by say $\varepsilon / \norm{g}_1$.

Let $Q(x)$ be the truncated series.
Then

$$
\abs{g\ast\phi_t(x) - g\ast Q(x)} \leq \abs{g\ast(\phi_t - Q)(x)} \leq \norm{g} \norm{p_t(x) - Q(x)}_\infty < \varepsilon \to 0,
$$

where $\norm{f}_\infty = \displaystyle\sup_{x\in [a, b]} \abs{f(x)}$ and $(g\ast Q)(x)$ is a polynomial. $\qed$

## Fourier Transform on $\RR^n$

Given $f\in L^1$, we defined the Fourier transform of $f$ by
$$
\hat{f}(\xi) = \int f(x) e^{-2\pi i x\cdot \xi}~dx
.$$

Some facts we know about the Fourier transform:

- $f\in L^1 \implies \hat{f}$ is bounded and uniformly continuous.
  
  (From an old homework!)

- The Riemann-Lebesgue lemma: $\displaystyle\lim_{\abs \xi \to \infty} \hat{f}(\xi) = 0$, i.e. $\hat{f}$ vanishes at infinity.

> Warning: it is **not** true that $f \in L^1 \implies \hat{f}\in L^1$!

### Fourier Inversion Formula

**Theorem (Inversion Formula):**
If $f, \hat{f} \in L^1$ then
$$
f(x) = \int \hat{f} (x) e^{2\pi i x\cdot \xi} ~d\xi \quad \text{for a.e. } x,
$$
i.e. $\hat{\hat{f}} = f(-x)$, and the Fourier transform is 4-periodic.

> Note that there is an interpretation here as writing an arbitrary function as a (continuous) sum of *characters*, where we're considering $\RR^n$ with the action of translation. 
In this setting, exponentials are certain eigenfunctions.

**Corollaries:**

1. $f, \hat{f} \in L^1$ implies that $f$ itself is bounded, continuous, and vanishes at infinity. (Note that this is not true for arbitrary $L^1$ functions!)

    > We will in fact show that $\theset{f \mid f, \hat{f} \in L^1}\injects L^1$ is dense.

2. $f \in L^1$ and $\hat{f} = 0$ a.e. $\implies f = 0$ almost everywhere 
   
   (Proof uses the Inversion formula)

*Proof of Inversion Formula:*

> Note: Fubini-Tonelli won't work here *directly*.

We'll have
$$
f(x) = \int\int f(y) e^{-2\pi i y\cdot \xi} e^{2\pi i x \cdot \xi} ~dy ~d\xi,
$$

which is (obviously?) not in $L^1(\RR^{2n})$.

So we'll introduce a "convergence factor" $e^{-\pi t^2 \abs{\xi}^2}$, which will make the integral swap result in something integrable, then take limits.

**Important example (HW):** 
$$
g(x) = e^{-\pi \abs{x}^2} \implies \hat{g}(\xi) = e^{-\pi \abs{\xi}^2}
.$$

Note that 
$$
g_t(x) = \frac{1}{t^n} e^{-\pi \abs{x}^2 / t^2}
$$ 
is an approximation to the identity, and $\int g_t = 1$.

By a HW exercise, have have 
$$
\hat{g_t}(\xi) = \hat{g}(t\xi) = e^{-\pi t^2 \abs{\xi}^2}
,$$ 
which is exactly the convergence factor we're looking for.
Moreover, $f \ast g_t \mapsvia{L^1} f$ in $L^1$.

> This says that the Fourier transform "commutes with dilation" in a certain way.

**Lemma (Multiplication Formula):**
If $f, g \in L^1$, then an easy application of Fubini-Tonelli yields
$$
\int f \hat{g} = \int \hat{f} g.
$$

We have

\begin{align*}
\int \hat{f}(\xi) e^{-\pi t^2 \abs{\xi}^2} e^{2\pi i x \cdot \xi} ~d\xi 
&\definedas 
\int \hat{f}(\xi) \phi(\xi) \quad \quad (= f \ast g_t(x) \mapsvia{L_1} f) \\
&= \int f(y) \hat{\phi}(y)~dy \\
&=_{DCT} \int \hat{f}(\xi) e^{2\pi i x \cdot \xi} ~d\xi 
\quad \text{as } t\to 0
.\end{align*}


where $\phi(\xi) = e^{2\pi i x\cdot \xi} \hat{g_t}(\xi)$.

> By a HW problem, we know 
$$
\hat{\phi}(y) = \hat{\hat{g}_t}(y-x) =  g_t(x - y)
.$$

But now one term is converging to $\int \hat{f}(\xi) e{2\pi i x\cdot \xi} ~d\xi$ as $t\to 0$ pointwise, and $f\ast g_t(x) \to f$ as $t\to 0$ in $L_1$.

So there is a subsequence of the latter term converging to $f$ almost everywhere, and thus the pointwise limit in the first is equal to the $L^1$ limit in the second.

We thus obtain 
$$
f(x) = \int \hat{f}(\xi) e^{2\pi i x\cdot \xi} ~d\xi
$$ almost everywhere.

$\qed$