# Tuesday October 22

## Convolution

Recall:

- **Continuous Compact Approximation:**
$C_c \injects L^1$ is dense.

- **Continuity in $L^1$:**  
  $$
  f\in L^1 \implies \norm{\tau f - f} \to 0, \text{ i.e. } \lim_{y\to 0} \int \abs{f(x+y) - f(x)} ~dx = 0
  .$$

- If $f\in L^1$, then for any $\varepsilon > 0$, 
  
  - **Small tails:** There exists a $\delta$ such that for all $E$ such that 
  $$
  m(E) \leq \delta \implies \int_E \abs{f} < \varepsilon
  .$$

  - There exists an $N$ such that 
  $$
  \int_{\theset{\norm{x} \geq N}} \abs{f} < \varepsilon
  .$$ 
    - Note that $\abs{f(x)} < \varepsilon ~~\forall x$ such that $\norm x \geq N$ exactly when $f$ is uniformly continuous

**Definition:**
The *convolution* of $f,g$ measurable functions on $\RR^n$ is given by
$$
f\star g (x) = \int_{\RR^n} f(x-y) g(y) ~dy
$$
for every $x$ for which this integral makes sense.


**Remarks:**

- There are sufficient conditions on $f,g$ which guarantee that $f\star g$ exists.

- If for some $x$, the function 
$$
y\mapsto f(x-y)g(y)
$$ is measurable, then the function
$$
y\mapsto f(y)g(x-y)
$$ 
is also integrable.

> Note that this is just a translation followed by a reflection, which is still integrable since this operation is in $\GL(n, \RR))$ and $f\star g = g \star f$.

## Properties of Convolutions

**Theorem 1:**

a. 
$$
f\in L^1 \text{ and } g \text{ bounded } \implies f\star g \quad \text{ is bounded  *and* uniformly continuous}.
$$
b. 
$$
f,g \in L^1 \text{ and } f,g \text{ bounded } \implies \lim_{\abs{x}\to\infty} (f\star g)(x) = 0
.$$

> Note that (b) immediately follows if it were the case that $f\star g$ were uniformly continuous and *integrable*, but we don't necessarily need integrability for this result.

> Note: It is possible to pointwise multiply 2 integrable functions and get something non-integrable -- consider $f^2$ where 
$$
f(x) = \frac{1}{\sqrt x} \chi_{[0,1]}
.$$

**Theorem 2:**

$$ f,g \in L^1 \implies \norm{f\star g}_1 \leq \norm{f}_1 \norm{g}_1,$$ and equality is attained if $f, g \geq 0$.

That is,
$$
\int \abs{f\star g} \leq \int\abs{f} \int\abs{g}
.$$

**Corollary**:
If $g$ is additionally *bounded*, then 
$$
\lim_{\abs x \to \infty} f\star g(x) = 0
.$$

**Theorem 3:**
\begin{align*}
f\in L^1, \quad g \text{ differentiable}, \quad \text{ and } g, \dd{g}{x_1}, \cdots, \dd{g}{x_n} &\text{ all bounded } \implies \\
f\star g \in C^1  
\text{ and }\dd{}{x_j} (f\star g) &= f\star (\dd{}{x_j} g)
.\end{align*}

**Corollary**:
$$
f\in L^1  \text{ and } g\in C^\infty_c \implies f\star g \in C^\infty \text{ and } \lim_{\abs x \to \infty} f\star g(x) = 0
.$$ 

In other words, defining $C_0$ as the functions that vanish at infinity, we have $f\star g \in C_0^\infty$.

> Note that we don't necessarily preserve *compact support* after this convolution. 
See the following picture, which looks similar for any fixed $x$ -- particularly any large $x$.

![Image](figures/2019-10-22-11:55.png)\

*Proof of Theorem 1, part (a):*

\begin{align*}
\abs{\int f(x-y)g(y) ~dy} 
&\leq \int \abs{f(x-y)} \abs{g(y)} ~dy \\
&\leq M \int \abs{f(x-y)} ~dy \\
&\leq M \norm{f}_1
.\end{align*}

and


\begin{align*}
\abs{f\star g(x+h) - f\star g}
&= \abs{
\int f(x+h-y)g(y)~dy - \int f(x-y)g(y)~dy
} \\
&\leq \int \abs{f(x+h-y) - f(x-y)} \abs{g(y)}~dy \\
&\leq M \int \abs{f(z+h) - f(z)}~dz \to 0
.\end{align*}


*Proof of Theorem 1, part (b):*

Let $\varepsilon > 0$, then choose $N$ such that 
$$
\int_{\theset{\norm{y} \geq N}} \abs{f(y)} ~dy < \varepsilon \quad \text{and} \quad 
\int_{\theset{\norm{y} \geq N}} \abs{g(y)} ~dy 
.$$

Since $\abs{x} \leq \abs{x-y} + \abs{y}$ by the triangle inequality, if we take $\abs{x} \geq 2N$, then *either* 

- $\abs{x-y} \geq N$, or
- $\abs{y} \geq N$.

In the first case, let $A_x = \theset{\abs x \geq N}$


\begin{align*}
\abs{f\star g} 
&\leq \int \abs{f(x-y)} \abs{g(y)} ~dy \\
&\leq M \int_{A_{x-y}} \abs{f(x-y)} < M\varepsilon
.\end{align*}


and in the second case, take


\begin{align*}
\abs{f\star g} 
&\leq \int \abs{f(x-y)} \abs{g(y)} ~dy \\
&\leq M \int_{A_{y}} \abs{g(y)} < M\varepsilon
.\end{align*}


*Proof of Theorem 2:*

Since $f,g \in L^1$, the function $h(x, y) \definedas f(x-y)g(y)$ will be measurable on $\RR^n\cross\RR^n$ as a product of measurable functions if we can show that the function $f_{x,y} \definedas (x,y) \mapsto f(x-y)$ is measurable.

To see that this is the case, define $F(x-y, y) = f(x-y)$ by taking the cylinder, then let 
$$
T = \left( \begin{array}{cc} 1& -1\\  0 & 1 \end{array}\right) \implies T(x, y) = (x-y, y)
,$$ 

Thus $f_{x,y}(x, y) = (F \circ T)(x, y)$.

We can now note that

\begin{align*}
\int \int \abs{f(x-y)} \abs{g(y)} ~dy ~dx 
&=_{FT} \int \int \abs{f(x-y)} \abs{g(y)} ~dx ~dy \\
&= \int \abs{g(y)} \left( \int \abs{f(x-y)} ~dx \right) ~dy \\
&= \norm{f}_1 \norm{g}_1
.\end{align*}

This proves that the integrand is in $L^1(\RR^{2n})$, so Fubini implies that $f\star g(x)$ is in $L^1$ for almost every $x$.

But then

\begin{align*}
\int \abs{f\star g(x)}~dx 
&\leq \int \int \abs{f(x-y) g(y)}~dy~dx \\
&= \norm{f}_1 \norm{g}_1
.\end{align*}

$\qed$

> Note that equality is attained here if $f, g \geq 0$.