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Consider the Linear Operator $T:L_2(\mathbb{R})\rightarrow L_2(\mathbb{R})$ defined by $$Tf(x)=\int_{\mathbb{R}}f(x-y)e^{-y^2}dy=f*e^{-y^2},\hspace{5pt}\forall f\in L_2(\mathbb{R})\text{ and }\forall x\in\mathbb{R}.$$ Using Young's Theorem, it is obvious that $\|Tf\|_2=\|f*e^{-y^2}\|_2\leq\|f\|_2\|e^{-y^2}\|_1$, so that $\|T\|\leq\|e^{-y^2}\|_1$. I just need to prove equality. I'm not sure what to pick as a particular $f$ (or sequence of $f_n$'s) to gain the equality. I was wondering if someone could give a proper proof. Thanks.

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It's easy to integrate Gaussians, so try using $f(x)=e^{-x^2/a^2}$. You should find $$ f * e^{-y^2}(x) = \sqrt{\pi}\frac{a}{\sqrt{1+a^2}} e^{-x^2/(1+a^2)}, $$ and then $$ \lVert f * e^{-y^2} \rVert_2^2 = \pi\frac{a^2}{\sqrt{1+a^2}} \sqrt{\frac{\pi}{2}}, $$ whereas $$ \lVert f \rVert_2^2 = a \sqrt{\frac{\pi}{2}}. $$ Since $$ \frac{a^2}{\sqrt{1+a^2}} \sim a $$ as $a \to \infty $, we conclude that the operator norm cannot be smaller than $\lVert e^{-y^2} \rVert_1 = \sqrt{\pi} $, after square-rooting the above equations.

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  • $\begingroup$ Thanks. Beautifully written. $\endgroup$ – Captain Jack Sparrow Oct 11 '15 at 16:16

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