Proof for $a\in (0,1)$ that $a\to F(a)= \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map, if $f\in L^p(X)$ for all $p\geq1$ in some measure space $(X,\mu)$.
I'd like to prove that for all $a_1,a_2\in(0,1)$ and $t\in[0,1]$ the following holds: \begin{align} F(t a_1+(1-t)a_2)\leq t F(a_1)+(1-t)F(a_2)\end{align} using the properties of the logarithm yields $F(a)=a \log\left(\int_X \lvert f\lvert^{1/a}\right)$. Plugging this in the above inequality reads \begin{align} (t a_1+(1-t)a_2)\log\left( \int_X \lvert f\lvert^{1/(t a_1+(1-t)a_2)}\right)\leq\\ t a_1 \log\left( \int_X \lvert f\lvert^{1/a_1}\right)+(1-t)a_2 \log\left( \int_X \lvert f\lvert^{1/a_2}\right)\end{align}
Is the $\log$ on the LHS dominated by those on the RHS? I don't know why? How can I continue? Apply Hölder?