# An inequality involving exponential of compact self-adjoint operator (SOLVED)

I've run into a tricky functional analysis problem. Here it is:

Suppose $$A: H \to H$$ is a compact self-adjoint operator on a Hilbert space H. Assume that the spectrum of $A$ is located in the open left-hand-side of the complex plane. Define the exponential of $A$ by $$\exp(A) = \sum_{n=0}^\infty \frac{A^n}{n!}\,.$$

Prove or disprove:

$$\|\exp(At)\| \leq M e^{-\alpha t}$$

in the usual operator norm for some $\alpha, M > 0$.

So far, I've shown that $\exp(A)$ is a bounded linear operator and that $$\exp(A) = \sum_{m=1}^\infty e^{\lambda_m}P_m$$ where $\{ \lambda_m\}$ are the eigenvalues of $A$ and $P_m$ is the orthogonal projection operator onto the corresponding eigenspace using the spectral theorem.

I think this inequality may be false, since when I attempted to derive a bound using the usual trick with the triangle inequality, then it didn't work out. I can't figure out how to disprove it though. I really hate these prove or disprove problems...

EDIT: Since $0$ is not in the spectrum H must be finite dimensional, and the problem is trivial.

-
The spectrum of $A$ contains $0$ unless $H$ is finite dimensional. – Jonas Meyer Dec 3 '12 at 6:33
Ah thanks so much! They don't have this theorem in my textbook. Now the problem is trivial. – B0112358 Dec 3 '12 at 7:01