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Let $$ A(f)(x) := \int_{0}^{x} f(t) \;dt $$

be defined on the Schwartz space $\mathcal{S}(\mathbb{R})$. Is $A$ a well-defined and bounded linear operator into $\mathcal{S}(\mathbb{R})$? If so, what is its operator norm?

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  • $\begingroup$ It doesn't map the Schwartz class to itself, first, and, second, since the Schwartz class is not a Banach space, there's no obvious natural "operator norm". $\endgroup$ – paul garrett Jun 11 '15 at 13:29
  • $\begingroup$ For which $f\in \mathcal S(\Bbb R)$ will we have $A(f)\notin\mathcal S(\Bbb R)$? $\endgroup$ – Berci Jun 11 '15 at 13:32
  • $\begingroup$ Can you prove the first statement? That's true, what I meant was the Lipschitz constant with respect to the metric, but that is irrelevant if it is not well-defined. $\endgroup$ – user159517 Jun 11 '15 at 13:35
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Taking $$f(t)=\frac{2}{\sqrt{\pi}}e^{-t^2}$$ results in $A(f)(x)=\text{erf}(x)$, the error function, which is obviously not a Schwartz function.

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