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Let $X$ be a Hilbert space and $A\in\mathcal{L}(X)$.

I want to show that $\displaystyle e^{At}:=\sum_{n=0}^{\infty}\frac{(At)^{n}}{n!}=T(t)$ defines a strongly continuous semigroup (i.e. a $C_{0}$-semigroup).

First I must show that $\forall t\ge 0$, $T(t)$ is a bounded linear operator on $X$, i.e. $T(t)\in\mathcal{L}(X)$.

I started by supposing that $T(t)$ is indeed in $\mathcal{L}(X)$.

Then define $\displaystyle T_{n}(t)=\sum_{k=0}^{n}\frac{(At)^{k}}{k!}$.

This is a Cauchy sequence, so

$\displaystyle ||T_{n}(t)-T_{m}(t)||_{X}=||\sum_{k=n}^{m}\frac{(At)^{k}}{k!}||_{X}\le\sum_{k=n}^{m}\frac{||A||_{X}^{k}t^{k}}{k!}\le\sum_{k=n}^{\infty}\frac{||A||_{X}^{k}t^{k}}{k!}\to0 $ as $n\to 0$

since the exponential power series $\displaystyle e^{||A||_{X}t}=\sum_{n=0}^{\infty}\frac{||A||_{X}^{n}t^{n}}{n!}$ converges.

$\mathcal{L}(X)$ is a Hilbert space, hence complete, so Cauchy sequences converge and thus we can define $T(t)$ as

$\displaystyle T(t):=\lim_{n\to\infty}T_{n}(t)=\sum_{n=0}^{\infty}\frac{(At)^{n}}{n!}$.

Does this prove that $T(t)$ is a bounded linear operator on $X$?

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yes, it does. L(X) is the space of bounded linear operators and it is an element of L(X) by definition.

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