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I was wondering why the schwarz functions $S(\mathbb{R})$ are dense inside the $L_p(\mathbb{R})$ spaces and I was reading this answer, but I don't understand why the $g_t$ are in $S(\mathbb{R})$. Could someone explain this?

roi_saumon
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1 Answers1

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Note the following first:

The convolution with any function with compact support is given by an integral on a compact interval.

So now you can establish the second point:

The convolution with any $C^\infty$ function given by an integral on a compact interval gives a $C^\infty$ function (basically you can take the derivatives inside).

John B
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  • The first point is clear, if I consider $supp(g)\subset [-R,R]$, then $f \ast g(x) =\int_{-R}^{R}f(x-t)g(t)dt$ but then why can I take the derivative inside, since I don't even know that $f$ is differentiable no? I just know it is in $L^p(\mathbb{R})$ right? – roi_saumon May 25 '20 at 10:45
  • You know that $fg=gf$. – John B May 25 '20 at 10:50
  • Okay, I get it now. Nice trick – roi_saumon May 25 '20 at 12:34