Controlling $C^1$ norm by Hölder norm and $C^0$ norm

Question

I've been trying to do this problem for a little while and have not made much progress. I want to show that for any $\epsilon>0$, there exists some $C_{\epsilon}>0$ such that $$ \Vert u'\Vert_{L^{\infty}} \le \epsilon\sup_{x\neq y}\frac{|u'(x)-u'(y)|}{|x-y|^{\alpha}}+C_{\epsilon}\Vert u\Vert_{L^{\infty}} $$ holds for all $u\in C^{1,\alpha}([0,1])$ where $0<\alpha<1$. I've looked at a few special cases, such as when the function is linear (in which case you can take $C_{\epsilon}=2$) or when the function is highly oscillatory, such as $\sin(nx)$ in which case you can take $C_{\epsilon}=1/\epsilon$, but I haven't had much success in the general case.

Answer 1

Suppose $$ |u'(x) - u'(y)| \le \frac1\epsilon |x-y|^\alpha \tag 1$$ Suppose $u'(x) \ge 1$. Then from (1), for $|y-x| \le (\frac12 \epsilon)^{1/\alpha}$, we have that $|u'(x) - u'(y)| \le \frac12$, and hence $u'(y) \ge \frac12$.

Hence if $|x-y| = (\frac12 \epsilon)^{1/\alpha}$, then $$ |u(x) - u(y)| = \left|\int_x^y u'(z) \, dz\right| \ge \tfrac12 (\tfrac12 \epsilon)^{1/\alpha} ,$$ thus ${\|u\|}_\infty \ge \frac14 (\frac12 \epsilon)^{1/\alpha}$. Similarly if $u'(x) \le -1$.

Thus if we set $C_\epsilon = 4 (2/\epsilon)^{1/\alpha}$, then $$ \epsilon\sup_{x\neq y}\frac{|u'(x)-u'(y)|}{|x-y|^{\alpha}}+C_{\epsilon}{\Vert u\Vert}_{L^{\infty}} \le 1 $$ implies that

$$ {\Vert u'\Vert}_{L^{\infty}} \le 1 . $$