Show that $\|e^{tA}\| \le e^{t\|\Re (A)\|}$

Question

Let $X$ be a complex Hilbert space, and let $A$ be a bounded linear operator on $X$. Define the real part of $A$ to be $\Re(A)=\frac{1}{2}(A^{\star}+A)$, and define $e^{tA}=\sum_{n=0}^{\infty}\frac{1}{n!}(tA)^{n}$, which converges in $\mathcal{L}(X)$ for all $t$. Show that $$ \|e^{tA}\| \le e^{t\|\Re(A)\|},\;\;\; t \ge 0. $$ Note: $A$ is not assumed to be normal.

Answer 1

Goes like that: let $v(t)=e^{At}u$. Then $$ \frac d{dt}||v(t)||^2=(Av(t),v(t))+(v(t),Av(t))=((A+A^*)v(t),v(t))\le ||A+A^*||||v(t)||^2, $$ whence $$ ||e^{At}u||^2\le e^{||A+A^*||t}||u||^2 $$ and, taking the square root, $$ ||e^{At}u||\le e^{\frac12||A+A^*||t}||u||. $$ QED