I am working on the following qual prep question.
Suppose $f:[0,1] \to \mathbb{R}$ is absolutely continuous and there exists $C>0$ such that $$\int_0^{1}e^{|f'(x)|}dx \leq C$$ Let $D=\{(x,y) \in [0,1] \times [0,1] : x \neq y\}$. Prove that $$\sup_{(x,y) \in D} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}} < \sup_{(x,y) \in D} (|x-y|^{1-\alpha}(|\log|y-x|| + C) )$$
I have a hint that we can apply Jensen's inequality to $$\frac{1}{x-y} \int_x^y|f'(t)|dt, \phi(t)=e^t$$ But I could not figure out how.
Jensen's inequality tells you that $$\phi \left( \frac 1{y-x} \int_x^y g(t) \,dt \right) \le \frac 1{y-x} \int_x^y \phi(g(t)) \, dt$$ whenever $g$ is integrable on $[0,1]$ because $\phi$ is convex. Since $f$ is absolutely continuous on $[0,1]$ its derivative is integrable and for any $x,y \in [0,1]$ the fundamental theorem of calculus gives you $$f(x) - f(y) = \int_x^y f'(t) \, dt.$$ Now let $x,y \in [0,1]$ and assume without loss of generality that $x < y$. Then $$\frac{|f(x) - f(y)|}{|x-y|} \le \frac 1{y-x}\int_x^y |f'(t)| \, dt$$ so that $$\exp\left(\frac{|f(x) - f(y)|}{|x-y|}\right) \le \exp\left(\frac 1{y-x}\int_x^y |f'(t)| \, dt\right) \le \frac 1{y-x} \int_x^y e^{|f'(t)|} \, dt \le \frac{C}{y-x} = \frac{C}{|x-y|}.$$ Now apply the logarithm to both sides to get $$\frac{|f(x) - f(y)|}{|x-y|} \le \log|x-y| + \log C$$ and $$\frac{|f(x) - f(y)|}{|x-y|^\alpha} \le |x-y|^{1-\alpha} \left( \log |x-y| + \log C \right).$$