- #1

joypav

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**Problem:**

Let $\phi(x), x \in \Bbb{R}$ be a bounded measurable function such that $\phi(x) = 0$ for $|x| \geq 1$ and $\int \phi = 1$. For $\epsilon > 0$, let $\phi_{\epsilon}(x) = \frac{1}{\epsilon}\phi \frac{x}{\epsilon}$. ($\phi_{\epsilon}$ is called an approximation to the identity.)

If $f \in L^1(\Bbb{R})$, show that

$lim_{\epsilon \rightarrow 0}(f ∗ \phi_{\epsilon})(x) = f(x)$,

for every Lebesgue point $x$.

The convolution $f ∗ g$ is defined as

$(f ∗ g)(x) = \int_{\Bbb{R}}f(y)g(y − x) dy$

Hint:

Note that $\int \phi_{\epsilon} = 1, \epsilon > 0$, so that

$(f ∗ \phi_{\epsilon})(x) − f(x) = \int[f(x − y) − f(x)] \phi_{\epsilon}(y) dy$

**Proof:**

Notice that, if $x \geq \epsilon \implies \phi_{\epsilon}(x)=\frac{1}{\epsilon} \phi(\frac{x}{\epsilon}) = \frac{1}{\epsilon} \cdot 0 = 0$ (because $\left| \frac{x}{\epsilon} \right| \geq 1$)

Let $u = \frac{x}{\epsilon}$.

Then we have,

$\int_{|x|<\epsilon} \phi_{\epsilon}(x) dx = \frac{1}{\epsilon^n} \int_{|x|<\epsilon} \phi(\frac{x}{\epsilon}) dx =$ (change of variable) $\int_{|u|<1} \phi(u) du = 1$ (by Hint)

$\implies \left| (f ∗ \phi_{\epsilon})(x) - f(x) \right| \leq \int_{|y|<\epsilon} \left| f(x-y) - f(x) \right| \cdot \left| \phi_{\epsilon}(y) \right| dy$

Now, we can bound $\phi_{\epsilon}(y)$ using that $\phi(x)$ is bounded...

meaning, $\phi(x)$ is bounded by some value, say $M>0$, so

$\phi_{\epsilon}(y) = \frac{1}{\epsilon^n} \phi(\frac{y}{\epsilon}) \leq \frac{M}{\epsilon^n}$

I don't know what step to take from here... is this the right idea?