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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.

daruma
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    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 08 '20 at 01:20
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    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

    https://math.stackexchange.com/questions/557061/convergence-of-a-integral-heat-kernel-and-dirac-delta-function

     – whpowell96 Jan 08 '20 at 01:30

1 Answers1

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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,

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

IW. Krisna Adipayana
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  • Okay, so I agree it is a convolution but how can I compute the limit? – daruma Jan 08 '20 at 01:15
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    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 08 '20 at 01:45
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    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 $$\lim\limits{n\to\infty}(fK_n)(x)=f(x)$$ whenever $ f $ is continuous at $ x $. Moreover, if $ f $ is continuous everywhere then $ fK_n \to f $ uniformly. – IW. Krisna Adipayana Jan 08 '20 at 02:01
  • Does this theorem have a name? – daruma Jan 08 '20 at 02:07
  • 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 08 '20 at 02:10
  • I have no idea about the name of this theorem, but you can found the equivalent theorem in Körner, T. W. (1989). Fourier analysis. p.254-255 (Theorem 51.5) for instance. – IW. Krisna Adipayana Jan 08 '20 at 02:19
  • If you want to see this geometrically, you can see a graph of $ g(x) $ in https://www.desmos.com/calculator/ekfuz4ivdl. – IW. Krisna Adipayana Jan 08 '20 at 02:40
  • Sorry, I made a little mistake $ g_n $ should be form $$ g_n(y)=\dfrac{1}{\pi}\dfrac{1/n}{y^2+(1/n)^2}. $$ – IW. Krisna Adipayana Jan 08 '20 at 02:59