"Approximation to the Identity" and "Convolution" Proof

[joypav]

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?

 

[Opalg]

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?

Since $|\phi_{\epsilon}(y)| \leqslant \frac M\epsilon$, the inequality $| (f ∗ \phi_{\epsilon})(x) - f(x) | \leqslant \int_{|y|<\epsilon} | f(x-y) - f(x) | \cdot | \phi_{\epsilon}(y) |\, dy$ tells you that $| (f ∗ \phi_{\epsilon})(x) - f(x) | \leqslant \frac M\epsilon \int_{|y|<\epsilon} | f(x-y) - f(x) |\, dy$. Now you need to compare that information with the definition of a Lebesgue point.

 

[math771]

Yes, you are on the right track. The next step would be to use the boundedness of $\phi(x)$ to bound the integral. Since $\phi(x)$ is bounded by $M$, we have:
$$\left|\int_{|y|<\epsilon} \left| f(x-y) - f(x) \right| \cdot \left| \phi_{\epsilon}(y) \right| dy\right| \leq \int_{|y|<\epsilon} \left| f(x-y) - f(x) \right| \cdot \left| \frac{M}{\epsilon^n} \right| dy$$
Now, since $f \in L^1(\Bbb{R})$, we know that it is integrable and therefore, there exists some $K>0$ such that $|f(x)| \leq K$ for all $x \in \Bbb{R}$. Therefore, we can further bound the integral as:
$$\int_{|y|<\epsilon} \left| f(x-y) - f(x) \right| \cdot \left| \frac{M}{\epsilon^n} \right| dy \leq \int_{|y|<\epsilon} \left| f(x-y) - f(x) \right| \cdot \left| \frac{K}{\epsilon^n} \right| dy$$
Now, using the definition of $f * \phi_{\epsilon}$, we can rewrite the integral as:
$$\int_{|y|<\epsilon} \left| f(x-y) - f(x) \right| \cdot \left| \frac{K}{\epsilon^n} \right| dy = \int_{\Bbb{R}} \left| f(x-y) - f(x) \right| \cdot \left| \frac{K}{\epsilon^n} \right| \phi_{\epsilon}(y) dy = \left|\left| f * \phi_{\epsilon} - f \right|\right|_1$$
where $|| \cdot ||_1$ denotes the $L^1$ norm. Now, since $f \in L^1(\Bbb{R})$, we know that $f * \phi_{\epsilon}$ converges to $f$ in the $L^1$ norm