# $T_{\epsilon}f(x)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\epsilon}{y^2+\epsilon^2}f(x-y) dy$

I am asked to show that for an $$L^1$$ function $$f$$,

$$T_{\epsilon}f(x)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\epsilon}{y^2+\epsilon^2}f(x-y) dy$$

converges to $$f(x)$$ as $$\epsilon\to 0^+$$ for almost every $$x$$.

As this question is given in the context of a measure theory course, I was thinking of using the dominated convergence theorem for the sequence of functions $$\displaystyle f_n(y)=\frac{1/n}{y^2+(1/n)^2}f(x-y)$$ for a fixed $$x$$.

Firstly $$f_n$$ is dominated by $$f_1\in L^1(\mathbb{R})$$ (i.e. $$|f_n|\leqslant |f_1|$$.)

Also $$f_n\to 0=:f$$. But then, $$\lim_{n\to\infty}\int_{-\infty}^{\infty}f_n(y) dy=\int_{-\infty}^{\infty}\lim_{n\to\infty}f_n(y)dy=\int 0 dy= 0$$

which is not the desired result.

Could someone explain what I reasoned wrongly and what is the correct approach.

Thank you.

• letting $y=\epsilon z$ then you get: $$T_{\epsilon}f(x)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{1}{z^2+1}f(x-\epsilon z)\,dz$$ which seems easier to deal with. – Thomas Andrews Jan 8 at 1:20
• I can't think of the details to show it for $L^1$ functions, but this kernel converges to the Dirac delta function, and most of those proofs work in the same way. Here is an example with a similar kernel (indexed by $t$ instead of $\epsilon$) on a Schwartz class. I believe you can more or less copy the structure of this proof but you won't be able to rely on continuity. I am pretty sure that the transform in Thomas's comment will be helpful too math.stackexchange.com/questions/557061/… – whpowell96 Jan 8 at 1:30

Use the convolution approach on $$L^1(\mathbb{R})$$,

$$(f*g)(x)=\int\limits_{\mathbb{R}}f(x-t)g(t)~\text{d}t=\int\limits_{\mathbb{R}}g(x-t)f(t)~\text{d}t.$$ Then, we can write

$$T_{\epsilon}f(x)=\dfrac{1}{\pi}\int\limits_{\mathbb{R}}f(x-y)g(y)~\text{d}y,$$

where,

$$g(y)=\dfrac{\epsilon}{y^2+\epsilon^2}.$$

• Okay, so I agree it is a convolution but how can I compute the limit? – daruma Jan 8 at 1:15
• Definition: Let $\{K_n(x)\}_{n=1}^{\infty}$ be a sequence of Functions such that $K_n:\mathbb{R}\to\mathbb{R}$. Then we call this a family of good Kernels of the following properties are satisfied: (i) For all $n\in\mathbb{N}$, $$\int\limits_{\mathbb{R}}K_{n}(x)~\text{d}x=1.$$ (ii) There exists $M>0$ such that for all $n\in\mathbb{N}$, $$\int\limits_{\mathbb{R}}|K_{n}(x)|~\text{d}x\leq M.$$ (iii) For every $\eta>0$, we have $$\int\limits_{|x|>\eta}|K_{n}(x)|~\text{d}x\to 0,$$ as $n\to \infty$. – IW. Krisna Adipayana Jan 8 at 1:45
• Then, we use the following theorem: If $\{K_n(x)\}_{n=1}^{\infty}$ ia a family of good kernels, $K_n:\mathbb{R}\to\mathbb{R}$, and $f: \mathbb{R}\to \mathbb{R}$ is integrable, then $$\lim\limits_{n\to\infty}(f*K_n)(x)=f(x)$$ whenever $f$ is continuous at $x$. Moreover, if $f$ is continuous everywhere then $f*K_n \to f$ uniformly. – IW. Krisna Adipayana Jan 8 at 2:01
• Does this theorem have a name? – daruma Jan 8 at 2:07
• Now, the problem to show that $T_{\epsilon}f(x)$ converges to $f(x)$ as $\epsilon \to 0^+$ for almost every $x$, is equal to show that $$g_n(y)=\dfrac{1/n}{y^2+(1/n)^2}$$ is a good kernel. Then use the above theorem to complete the proof. But, your $f$ is must be continuous. – IW. Krisna Adipayana Jan 8 at 2:10