$a\mapsto \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map 
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?
 A: Let $a_1, a_2\in (0,1)$ and $t\in(0,1)$, then by Hölder inequality with 
$$
p=\frac{ta_1+(1-t)a_2}{ta_1}\quad\text{ and }\quad q=\frac{ta_1+(1-t)a_2}{(1-t)a_2}
$$
we get
$$
\int\limits_X |f|^{\frac{1}{ta_1+(1-t)a_2}}d\mu=
\int\limits_X |f|^{\frac{t}{ta_1+(1-t)a_2}}|f|^{\frac{1-t}{ta_1+(1-t)a_2}}d\mu
\leq
\left(\int\limits_X \left(|f|^{\frac{t}{ta_1+(1-t)a_2}}\right)^{\frac{ta_1+(1-t)a_2}{ta_1}}d\mu\right)^{\frac{ta_1}{ta_1+(1-t)a_2}}\left(\int\limits_X \left(|f|^{\frac{1-t}{ta_1+(1-t)a_2}}\right)^{\frac{ta_1+(1-t)a_2}{(1-t)a_2}}d\mu\right)^{\frac{(1-t)a_2}{ta_1+(1-t)a_2}}=
\left(\int\limits_X |f|^{\frac{1}{a_1}}d\mu\right)^{\frac{ta_1}{ta_1+(1-t)a_2}}\left(\int\limits_X |f|^{\frac{1}{a_2}}d\mu\right)^{\frac{(1-t)a_2}{ta_1+(1-t)a_2}}
$$
Hence
$$
\left(\int\limits_X |f|^{\frac{1}{ta_1+(1-t)a_2}}d\mu\right)^{ta_1+(1-t)a_2}\leq
\left(\int\limits_X |f|^{\frac{1}{a_1}}d\mu\right)^{ta_1}\left(\int\limits_X |f|^{\frac{1}{a_2}}d\mu\right)^{(1-t)a_2}
$$
After taking lorarithms we see that the this inequality is equivalent to 
$$
F(t a_1+(1-t)a_2)\leq tF(a_1)+(1-t)F(a_2)
$$
Hence $F$ is convex.
