Semigroups: given $\|S_T f\| \leq C$, what can I say about $\|S_t f\|$ for $t \leq T$?

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Assume that we have a strongly continuous semigroup $(S_t)_{t \geq 0}$ of linear operator on a Banach $X$ such that for all $f \in X$, $$\|S_T f\| \leq C \|f\|, $$ for some constant $C > 1$ and some fix $T > 0$. Can I deduce that for all $t \in (0,T)$, $$ \|S_t f\| \leq C \|f\| ? $$

Ok so I did some thinking and also found this post Is the Norm of the Square Root of an Operator equal to the Square root of the Norm of the Operator, but I'm not sure that the spectral theorem applies with such general hypothesis. Can anyone confirm ?

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If $(S(t))_{t \ge 0}$ is a $C_0$-semigroup in a Banach space $X$, then, there exist $M>1$ and $\omega \ge 0$ such that $$\|S(t)\|\le M e^{\omega t}, \quad \text{ for all } t\ge 0.$$ In particular, $$\|S(t)\|\le M e^{\omega T}, \quad \text{ for all } 0\le t\le T.$$ The constant $C=M e^{\omega T}>1$ is valid for $S(T)$ and also $S(t)$, $0\le t\le T$.

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Let $\mathcal{H}$ be a Hilbert space. If $T : \mathcal{H} \to \mathcal{H}$ is a bounded linear operator, then its operator norm is given by

$$ \|T\| = \sup_{\substack{x \in \mathcal{H} \\ x \neq 0}} \frac{\|T x\|}{\|x\|} = \sup_{\substack{x \in \mathcal{H} \\ x \neq 0}} \sqrt{\frac{\langle x, T^{\mathsf{T}}T x\rangle}{\langle x, x\rangle}}. $$

In the special case $\mathcal{H} = \mathbb{R}^d$, then $\|T\|$ is the square root of the maximum eigenvalue of the positive semi-definite symmetric matrix $T^{\mathsf{T}}T$. Then your question can be rephrases as

Question. If $(S_t)_{t\geq 0}$ is a strongly continuous semi-group on a Banach space $X$, then is the map $t \mapsto \|S_t\|$ non-decreasing in $t$?

Consider $X=\mathbb{R}^3$ with $S_t = e^{tA}$, where

$$ A = \begin{pmatrix} 0 & 1 & -4 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}. $$

Then using numerical computation, we find that

\begin{align*} \| S_5 \| &= \left\| \begin{pmatrix} 1 & 5 & -\frac{15}{2} \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{pmatrix} \right\| = 97.9222\cdots \end{align*}

while

\begin{align*} \|S_7\| &= \left\| \begin{pmatrix} 1 & 7 & -\frac{7}{2} \\ 0 & 1 & 7 \\ 0 & 0 & 1 \end{pmatrix} \right\| = 75.3107\cdots. \end{align*}