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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?

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up vote 2 down vote accepted

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.

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