19

Let $X$ be a Banach space. $A\in B(X)$ is a bounded operator. we can define $e^{tA}$ by

$$e^{tA}=\sum_{k=0}^{+\infty}\frac{t^kA^k}{k!}$$ I am interested in this property:

If $x\in X$, such that the function $t\mapsto e^{tA}x$ is bounded on $\mathbb{R}$, then we have necessarily $$\inf_{t\in \mathbb{R}}|e^{tA}x|>0 \ \ \ \ \ or \ \ \ \ e^{tA}x=0 \ \ (i.e. \ \ x=0).$$

This property is clear in the scalar case $A=a\in \mathbb{C}$. Because $t\mapsto e^{ta}x$ is bounded on $\mathbb{R}$ if and only if $Re(a)=0$, and then $\inf_{t\in \mathbb{R}}|e^{ta}x|=|x|$.

This property is also true if $X$ is finite dimensional i.e if $A$ is a matrix, and was answered here in math stack exchange.

So my question is: "Does this property hold if $X$ is infinite dimensional ? "

user165633
  • 2,721
  • 15
  • 34

1 Answers1

5

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 $t\to\pm\infty$.

fedja
  • 15,893
  • 1
  • 27
  • 40