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This is problem 3.12 in Paul Garrett's 2019 example problems.

For $f \in \mathscr{S} $ (the space of Schwartz functions), show that $$\lim_{t \rightarrow + \infty} f(x) * \frac{2 \sin tx}{tx} = f(x) $$ ($*$ is the convolution operator).

I'm not sure in what sense the expression is supposed to converge (pointwise?).

I'd appreciate a hint about where to start; My first thought was to try showing that $2 \operatorname{sinc}(tx)$ are approximate identities, but their integrals are all $2 \pi$, so this approach quickly seemed like it might not work.

Michael Hardy
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thisre
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  • You say the asterisk represents the convolution operator. Does this mean convolution as a function of $t,$ or as a function of $x$? $\qquad$ – Michael Hardy Nov 14 '19 at 05:16
  • ok, I think it must mean as a function of $x,$ but this notation seems abominable. – Michael Hardy Nov 14 '19 at 05:17
  • I see: Garrett himself uses this notation. – Michael Hardy Nov 14 '19 at 05:19
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    Apologies for the lack of response. Yes, convolution as a function of $x$, so the expression is $\int_{\mathbb{R}} f(x-y) \frac{2 \sin ty}{ty} dy.$ – thisre Nov 14 '19 at 05:39
  • $$ \begin{align} & \frac 1 t \int\limits_{\mathbb R} f(x-y) \frac{2\sin(ty)}{ty} \big( t, dy\big) \ {} \ = {} & \frac 1 t \int\limits_{\mathbb R} f\left( x - \frac u t \right) \frac{2\sin u} u , du \ {} \ \approx {} & \frac 2 t \int_{-1/t}^{+1/t} f\left( x-\frac u t \right) , du \text{ ?} \end{align} $$ But ignore the above, unless there is something in it worth not ignoring. – Michael Hardy Nov 14 '19 at 06:32

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