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I am trying to prove the following statement.

if $f\in L^2$ and $g\in L^1$, $\mathcal{F}$ denotes fourier transformation,and$*$ denotes convolution, then $$\mathcal{F}(f*g)=\mathcal{F}(f)\mathcal{F}(g)$$

It should be extended from the case $f,g\in L^1$.

I followed the procedure starting from the lower part of page 72, but I couldn't understand the last estimation

$$\int\int|f(x-y)-f_n(x-y)||g(y)|\,dx\,dy\\\le||f-f_n||_2||g||_1$$

Any hints? Thanks!

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1 Answer 1

up vote 1 down vote accepted

I think there might be a typo in how that inequality is written down. To show the $L_2$ limit of $\mathcal{F}(f_n∗g)$ is $\mathcal{F}(f∗g)$:

$$ ||f_n*g - f*g||_2 = ||(f_n-f)*g||_2 \leq ||f_n-f||_2 ||g||_1\to 0 $$

I guess this is what the written solution was going for.

Context:

We want to show $\mathcal{F}(f*g)=\mathcal{F}(f)\mathcal{F}(g)$ You know there's a sequence $f_n\in L_1\cap L_2$ tending to $f$ in the $L_2$ norm. You know that for each $n$, $$\mathcal{F}(f_n*g)=\mathcal{F}(f_n)\mathcal{F}(g)$$ pointwise a.e. By (b) we know that the ($L_2$ and hence) pointwise a.e. limit of $\mathcal{F}(f_n*g)$ is $\mathcal{F}(f*g)$.

Therefore it suffices to prove that the pointwise a.e. limit of $\mathcal{F}(f_n)\mathcal{F}(g)$ is $\mathcal{F}(f)\mathcal{F}(g)$. But $\mathcal{F}(f_n)$ tends to $\mathcal{F}(f)$ in $L_2$ (Plancherel's theorem) and hence pointwise a.e.

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Thanks for the reply. My main difficulty is the estimation I stated above. And why $\mathcal{F}(g)\in L_\infty$? –  newbie May 2 '12 at 10:15
    
$|\mathcal{F}(g)(\zeta)|\leq ||g||_1$, but actually I don't think you even need that - I've modified my answer to give a full argument. –  Colin McQuillan May 2 '12 at 15:40

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