I have to choose whether the following operator

$$f\in L^2(\mathbb{R})\mapsto\int_{\mathbb{R}}f(x) e^{-x^2} dx$$

is a compact operator. I have tried to use the Cauchy Schwarz inequality

$$\langle f, e^{-x^2}\rangle \leq \|f\|_{L^2} \cdot \underbrace{\|e^{-x^2}\|_{L^2}}_{\leq\pi} \leq\pi\cdot\|f\|_{L^2} $$

but how can I identify the compactness of an operator?

  • 2
    $\begingroup$ Any continuous linear map into a finite dimensional vector space is necessarily compact. $\endgroup$ – Omnomnomnom Nov 20 '18 at 15:36

$T:X\to Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2\to\mathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $\mathbb R$ are relatively compact, so the operator is compact.

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