Is $e^{\gamma t}$ Hölder continuous, with $\gamma<0$. Is  $e^{\gamma t}$ Hölder continuous?, with $\gamma<0$.
This question appear in something that i am working, i am not sure if the answer is yes or not.
My only attempt is for definition $|e^{\gamma t}-e^{\gamma s}|=|\int_{s}^{t}\frac{1}{\gamma}e^{\gamma x}dx |\leq \frac{1}{|\gamma||}|t-s|$. If somebody knows can help me please, thank you so much.

Yes. Notice that since $\gamma < 0$, $0\leq e^{\gamma t} < 1$, hence $|e^{\gamma s}-e^{\gamma t}| \leq 1$. Now, using your computation, for any $\alpha\in[0,1]$,
$$
|e^{\gamma s}-e^{\gamma t}| = |e^{\gamma s}-e^{\gamma t}|^{\alpha}\, |e^{\gamma s}-e^{\gamma t}|^{1-\alpha} \leq \left(\tfrac{1}{|\gamma|}\,|t-s|\right)^\alpha 1^{1-\alpha}.
$$
Therefore
$$
\frac{|e^{\gamma s}-e^{\gamma t}|}{|t-s|^\alpha} \leq \tfrac{1}{|\gamma|^\alpha}.
$$
