For $f=\sin(x)$ I want to prove that $T(t)f$ does not converge uniformly to $f\in\mathbb{R}$ as $t\to 0$. Note that $T(t)$ is the Ornstein-Uhlenbeck semigroup.
Now, I have a theorem which states that for $f\in BUC(X)$, we have that $\lim_{t\downarrow 0}\|T(t)f-f\|_{\infty}=0\iff\lim_{t\downarrow 0}\|f(e^{-t}\cdot)-f\|_{\infty}=0$.
Observe that for $f(x)=\sin(x)$ we have
$$\lim_{t\downarrow 0}\|\sin(e^{-t}x)-\sin(x)\|_{\infty}=\|\sin(x)-\sin(x)\|_{\infty}=0.$$
Therefore $\lim_{t\downarrow 0}\|T(t)f-f\|_{\infty}=0$. However, this isn't the contradiction that I wanted. In order to prove uniform convergence of $T(t)f\to f$ as $t\to 0$ do I also have to prove it for $t\uparrow 0$? But in this case the theorem I used will no longer be applicable.