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.

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    $\begingroup$ 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. $\endgroup$ – Thomas Andrews Jan 8 at 1:20
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    $\begingroup$ 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/… $\endgroup$ – 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

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


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

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  • $\begingroup$ Okay, so I agree it is a convolution but how can I compute the limit? $\endgroup$ – daruma Jan 8 at 1:15
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    $\begingroup$ 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 $. $\endgroup$ – IW. Krisna Adipayana Jan 8 at 1:45
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    $\begingroup$ 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. $\endgroup$ – IW. Krisna Adipayana Jan 8 at 2:01
  • $\begingroup$ Does this theorem have a name? $\endgroup$ – daruma Jan 8 at 2:07
  • $\begingroup$ 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. $\endgroup$ – IW. Krisna Adipayana Jan 8 at 2:10

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