Prove an interpolation inequality

Solution 1:

Set $t\in (0,1)$ such that $(1-t)\beta + t = \gamma$, then we have $$ \frac{|u(x)-u(y)|}{|x-y|^\gamma}= \left( \frac{|u(x)-u(y)|}{|x-y|^{\beta}} \right)^{1-t}\left( \frac{|u(x)-u(y)|}{|x-y|}\right)^t\leq [u]_{\beta}^{1-t}[u]_{1}^t. $$ Also, we always have $| u|=|u|^{1-t}|u|^t$. Combining this with the previous estimate we get $$ \| u\|_\gamma \leq |u|_\infty^{1-t}|u|_\infty^t + [u]_{\beta}^{1-t}[u]_{1}^t =:a_1^{1-t}b_1^t+a_2^{1-t}b_2^t. $$ Now just write $A=a_1+a_2$, $A_i=a_i/A$, then the RHS becomes $$ A^{1-t}\left( A_1\left( \frac{b_1}{A_1}\right)^t + A_2 \left( \frac{b_2}{A_2}\right)^t \right) \leq A^{1-t}(b_1+b_2)^t, $$ where the last inequality is the concavity of the function $s\mapsto s^t$, since $A_1+A_2=1$. Combining all this gives the desired inequality.