# "Approximation to the Identity" and "Convolution" Proof

• MHB
• joypav
In summary: A Lebesgue point is a point where the derivative with respect to the variable at that point is zero. So, since $|\phi_{\epsilon}(y)| \leqslant \frac M\epsilon$, $| (f ∗ \phi_{\epsilon})(x) - f(x) | \leqslant \frac M\epsilon \int_{|y|<\epsilon} | f(x-y) - f(x) |\, dy$. This means that the Lebesgue point at which $f(x) = f(x)$ is at $x = \frac{1}{M}$.
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?

joypav said:
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.

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

## 1. What is the purpose of using "Approximation to the Identity" in a proof?

The purpose of using "Approximation to the Identity" in a proof is to show that a function can be approximated by a series of simpler functions that converge to the original function. This allows for easier manipulation and analysis of the function.

## 2. How does "Convolution" play a role in this proof?

"Convolution" is used in the proof to show that the approximating functions can be combined in a specific way to get closer to the original function. This is done by taking the integral of the product of the two functions, which results in a new function that is closer to the original.

## 3. What are the key steps in a proof involving "Approximation to the Identity" and "Convolution"?

The key steps in a proof involving "Approximation to the Identity" and "Convolution" include defining the approximating functions, showing that they converge to the original function, using convolution to combine the functions, and then showing that the resulting function is a good approximation of the original.

## 4. Can "Approximation to the Identity" and "Convolution" be used for all types of functions?

Yes, "Approximation to the Identity" and "Convolution" can be used for a wide range of functions, including continuous, discontinuous, and even periodic functions. However, the specific techniques and methods used may vary depending on the type of function being approximated.

## 5. Are there any limitations to using "Approximation to the Identity" and "Convolution" in a proof?

One limitation of using "Approximation to the Identity" and "Convolution" in a proof is that it may not always result in an exact solution. In some cases, the approximating functions may not be able to perfectly capture the behavior of the original function. Additionally, the process of combining the functions using convolution can be complex and may not always be feasible for certain types of functions.

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