How to prove that $f(t)=t\sin (\frac{1}{t})$ is 1-Hölder continuous (Lipschitz continuous) for fixed $t_0 \in [-1,1]$? $f(t)$ is defined as $0$ at $t_0 =0$.
That is, for $t_0 \in [-1,1]$, there exists a constant $C(t_0)\gt 0$ such that $$|f(t_0 +h)-f(t_0)| \le C(t_0)|h|, \forall t_0 +h\in [-1,1].$$ For $t_0=0$, it's easy to verify.
For $t_0\ne 0$, since $|f'(t_0)|\lt \infty$, so, $\forall \epsilon \gt 0$, there exists $\delta (t_0,\epsilon)\gt 0$, such that $$|f(t_0 +h)-f(t_0)| \le (|f'(t_0)|+\epsilon )|h|, \forall |h|\le \delta (t_0,\epsilon).$$ But I don't figure out how to deal with the case $|h|\gt \delta (t_0,\epsilon)$.
What's more, I also have no idea about a similar question:
$f(t)=t\sin (\frac{1}{t})$ is $\alpha$-Hölder continuous in [-1,1] for $\alpha \le \frac{1}{2}$, i.e. there exists a constant $C(\alpha) \gt 0$, such that $$|f(t_1)-f(t_2)|\le C(\alpha)|t_1-t_2|^{\alpha}, \forall t1, t2 \in [-1,1].$$ Can anyone help me?
Hints:
Given that you've dealt with what happens at 0, you essentially need to prove that a continuously differentiable function on a compact set is Lipschitz. It is crucial that $|f'(x)|$ is uniformly bounded for $x\in(-1,0)\cup(0,1)$. By the mean value theorem $|f(x)-f(y)|\leq ||f'||_\infty|x-y|$ for $x,y\in(-1,0)$, and also for $x,y\in(0,1)$. If you have that the function is Lipschitz on a finite number of overlapping intervals you can use the triangle inequality to get it on the whole interval: $|f(x)-f(z)+f(z)-f(y)|\leq C|x-z|+C|z-y| \leq 2C|x-y|$. (Actually you haven't shown that it's Lipschitz on an interval around 0, but if you carefully consider the different possibilities of $t_0$ and $t_0+h$, you'll see that you have all of the cases covered)
For $0<h,\alpha<1$, we have $|h|<|h|^\alpha$. For $h<2$ that means $|h|=2|h/2|<2|h/2|^\alpha=2^{1-\alpha}|h|^\alpha$.