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Let $X$ be the Banach space $X:=\{ f\in C(\mathbb{R},\mathbb{R}),\sup_{t\in \mathbb{R}}|e^{-s^2}f(s)|<+\infty \}$ equipped with the norm $$|f|_X=\sup_{t\in \mathbb{R}}|e^{-s^2}f(s)|$$

I want to show that the operator $A_a$ defined on $X$ by

$$(A_a f)(s)=f(a+s)$$ is a bounded operator on $X$. I get stuck in finding a constant $M$ such that $$|A_af|_X=\sup_{s\in\mathbb{R}}|e^{-s^2}f(a+s)|\leq M\sup_{s\in\mathbb{R}}|e^{-s^2}f(s)|=M|f|_X.$$ Also can we replace the function $e^{-s^2}$ by any function $\rho(s)$ with $0$ as limit in both infinities ?

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I found that the boundedness of the operator $A_\alpha$ depends on the function $\rho$. For $\rho(s)=e^{-s^2}$, the operator $A_\alpha$ is unlikely to be bounded, but we can make $A_\alpha$ bounded by assuming this condition on the function $\rho$: $$\frac{\rho(s-\alpha)}{\rho(s)}\leq M_\alpha, \ \ \forall s\in \mathbb{R}.$$ We will get then $$|A_af|_X=\sup_{s\in\mathbb{R}}|\rho(s)f(a+s)|=\sup_{s\in\mathbb{R}}|\rho(s-a)f(s)|\leq M_\alpha\sup_{s\in\mathbb{R}}|\rho(s)f(s)|=M_\alpha|f|_X.$$

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If you take $f(s)=e^{s^2} \in X$ you can see that its not bounded.

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Yes, I think even for $\rho(s)=e^{-s^2}$, it is not bounded. –  user144542 May 20 at 18:14

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