Convergence of the exponential for a densely-defined linear operator on a Hilbert space

Question

Let

• $H$ be a $\mathbb R$-Hilbert space
• $D(A)$ be a dense subspace of $H$ and $A$ be a linear operator from $D(A)$ to $H$
• $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ with $$-Ae_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_{n+1}\ge\lambda_n\;\;\;\text{for all }n\in\mathbb N\tag 2$$

Let $t\ge 0$. Is $$e^{-tA}x:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle x,e_n\rangle_He_n\tag 3$$ a well-defined bounded linear operator on $H$?

The actual question is: Is the series in $(3)$ convergent? By the comparison test, it's sufficient to show that $\sum_{n\in\mathbb N}e^{-\lambda_nt}$ is convergent.

Let $$M:=\sup_{n\in\mathbb N}\lambda_n\in(0,\infty]$$ and $$\alpha_n:=\left|\frac{e^{-\lambda_{n+1}t}}{e^{-\lambda_nt}}\right|=e^{-(\lambda_{n+1}-\lambda_n)t}\;\;\;\text{for }n\in\mathbb N\;.$$ By the ratio test, it's sufficient to show that $$\alpha_n\le q\;\;\;\text{for all }n\in\mathbb N\tag 4$$ for some $q\in[0,1)$. However, by $(2)$ we only obtain $\alpha_n\le 1$ for all $n\in\mathbb N$. Moreover, if $M<\infty$, then $$\lambda_{n+1}=\lambda_n\;\;\;\text{for all }n\ge N\tag 5$$ for some $N\in\mathbb N$ and hence $\sum_{n\in\mathbb N}e^{-\lambda_nt}$ should be divergent.

So, do we need to assume $M=\infty$? And how do we prove the convergence in that case?

Answer 1

Let $\lambda_n$ be a monotone increasing sequence, for every $x\in H$ you have $\langle x, e_n\rangle$ is square summable. Since $\lambda_n$ is monotone increasing you have $|e^{-t\lambda_n}|≤|e^{-t\lambda_0}|$ and for that reason $e^{-t \lambda_n}\langle x, e_n\rangle$ is square summable, being majorised by the square summable $e^{-t\lambda_0 } |\langle x, e_n\rangle|$.

For this reason the map $$e^{-tA}: H\to H, \qquad x\mapsto \sum_n e^{-t \lambda_n} \langle x,e_n\rangle e_n$$ is well defined. Linearity is trivial also.

Boundedness of the $\lambda_n$ is not necessary, you only need that the operator be bounded from below.

Note that if $A$ is bounded on $D(A)$ it extends to a bounded operator on $H$ and also that $e^{-t A}:=\sum_n (-tA)^n/n!$ is well defined and converges in operator norm.

Answer 2

Because your set $\{ e_n \}$ is orthonormal, then $\sum_{n}\alpha_n e_n$ converges in $H$ iff $\sum_{n}|\alpha_n|^2 < \infty$. In this case, $$ \|\sum_{n} e^{-\lambda_n t}\langle x,e_n\rangle e_n\|^2 = \sum_{n}e^{-2\lambda_n t}|\langle x,e_n\rangle|^2 \le \sum_{n}|\langle x,e_n\rangle|^2=\|x\|^2. $$ So your operator $e^{-tA}$ is a bounded operator with $\|e^{-tA}\| \le 1$ for all $t > 0$. In fact, $\|e^{-tA}\| \le e^{-\lambda_1 t}$. And, you do not need to assume that the $\lambda_n \le M$ for all $n$ and some $M$.