A property of exponential of operators

Solution 1:

The answer is "No". Take $C_0^\infty(\mathbb R)$ with the norm $\|f\|=\int_{\mathbb R}\frac{1}{1+y^2}|\widehat f(y)|\,dy$. Let $\psi$ be any $C_0^\infty(\mathbb R)$-function such that $\psi(x)=x$ on $[-1,1$. Let $Af=i\psi f$. Then we can check the boundedness of $A$ on the Fourier side by just checking that the convolution with $\widehat\psi$ is bounded in the weighted $L^1$ with the weight $\frac 1{1+y^2}$, i.e., by checking that $$ \int_{\mathbb R}\frac 1{1+(y+z)^2}|\widehat\psi(z)|\le \frac C{1+y^2}\,, $$ which is not difficult because $\widehat\psi$ decays faster than any power ($\widehat\psi(z)=O(z^{-2})$ as $z\to\infty$ is already enough to establish this claim). To turn it into an operator acting on a Banach space, just take the completion and extend by continuity.

Now take any smooth $g$ supported on $[-1,1]$. Then $(e^{tA}g)(x)=g(x)e^{itx}$, so we get a pure shift on the Fourier side. However, the integral of $| \widehat g |$ is finite and the shift carries carries any compact set away to infinity where the weight $\frac 1{1+y^2}$ is almost $0$. Thus $\|e^{tA}g\|\to 0$ as $t\to\pm\infty$.