In Tao's proof of Hölder’s inequality 
(Hölder’s inequality) Let $f \in L^p$ and $g \in L^q$ for some $0 < p,q \leq \infty$. Then $fg \in L^r$ and $\|fg\|_{L^r} \leq \|f\|_{L^p} \|g\|_{L^q}$, where the exponent $r$ is defined by the formula $\frac{1}{r} = \frac{1}{p} + \frac{1}{q}$.

In his lecture note, Tao reduces the problem to the nontrivial essential case where $p,q<\infty$, $r=1$, and  $\|f\|_{L^p}=\|g\|_{L^q} = 1$. The following is his proof:

Our task is now to show that
  $$\int_X |fg|\ d\mu \leq 1. \tag{*}$$
Here, we use the convexity of the exponential function $t \mapsto e^t$ on ${}[0,+\infty)$, which implies the convexity of the function $t \mapsto |f(x)|^{p(1-t)} |g(x)|^{qt}$ for $t \in [0,1]$ for any $x$. In particular we have
  $$
|f(x) g(x)| \leq \frac{1}{p} |f(x)|^p + \frac{1}{q} |g(x)|^q \tag{**}
$$
  and the claim $(*)$ follows from the normalisations on $p, q, f, g.$

Here are my questions:


*

*How is the convexity of the function $t \mapsto |f(x)|^{p(1-t)} |g(x)|^{qt}$ for $t \in [0,1]$ for any $x$ built up?

*How is the convexity used to get $(**)$?

 A: It is routine execrise to show that
$$
t\rightarrow e^{t}
$$
is a convex function by checking second derivative. Now define the function (for fixed $x)$
$$
t\rightarrow p(1-t)\log(|f(x)|)+qt\log(|g(x)|)\rightarrow e^{p(1-t)\log(|f(x)|)+qt\log(|g(x)|)}
$$
and it is the composition of a linear function with a convex function. Thus it must be convex as well. 
To get the desired inequality, now it suffice to set $t=\frac{1}{q}$. Recall that $p,q$ are Holder conjugates, so you have $\frac{1}{p}+\frac{1}{q}=1$. Recall that a function is convex implies $$F(t)\le (1-t)*F(0)+tF(1)$$ 
Thus setting $t=1/q$ we have 
$$
|f(x)g(x)|\le 1/q*|g(x)|^{q}+\frac{1}{p}*|f(x)|^{p}
$$
and we recover the original inequality. Here 
$$
t=1/q,F=e^{p(1-t)\log(|f(x)|)+qt\log(|g(x)|)}, t\in [0,1]
$$
As a side remark, it is well known that Holder's inequality can be viewed as a form Jensen's inequality, but Tao's proof is still very nice for pedalogical purposes. 
A: I'm not familiar with Tao's proof but here's the proof of Holder's inequality I know in the non-trivial case. It does not constitute a proper answer to your question but I believe it can highlight the use of convexity of $\exp(\cdot)$.

Set $$ F := \frac{|f|}{\|f\|}~~~~~~~~~~\text{and}~~~~~~~~~~ G :=
 \frac{|g|}{\|g\|}.$$ You can easily check that $$ \int_X F^p \; d\mu ~=~
 \int_X G^q \; d\mu ~=~ 1.$$ If $x \in X$ is such that $0 < F(x), G(x)
 < \infty$, then there exists numbers $s$ and $t$ such that $$ F(x) =
 \exp\left(\frac{s}{p}\right)~~~~~~~~~~\text{and}~~~~~~~~~~ G(x) = \exp\left(\frac{t}{q}\right).$$
  Since $p$ and $q$ are conjugate exponents, the convexity of
  $\exp(\cdot)$ gives $$ F(x)G(x) ~=~ 
 \exp\left( \frac{1}{p} s + \frac{1}{q} t\right) \\=~ \exp\left(
 \frac{1}{p} s + \left(1-\frac{1}{p}\right) t\right) \\\leq~
 \frac{1}{p}\exp(s) + \left(1 - \frac{1}{p}\right) \exp(t) \\=~ 
 \frac{1}{p}\exp(s) +  \frac{1}{q} \exp(t)  \\=~ \frac{1}{p}F(x)^p + \frac{1}{q}G(x)^q$$ for all $x
 \in X$. Now integrate over $X$ to get $$ \int_X FG\; d\mu ~\leq~
 \frac{1}{p} + \frac{1}{q} ~=~ 1$$ and finaly $$\int_X |fg| \; d\mu
 ~\leq~ \left(\int_X |f|^p\; d\mu\right)^\frac{1}{p}\left(\int_X |g|^q \;
 d\mu\right)^\frac{1}{q}.$$

A: Let $A=|f(x)|$ and $B=|g(x)|$. 
I just found that one does not need the convexity of the map
$$
t\mapsto A^{p(1-t)}B^{qt}
$$
and thus
the first question is unnecessary. By the convexity of the exponential function, 
$$
A^{p(1-t)}B^{qt}=\exp[(1-t)p\log A+tq\log B]
\leq (1-t)A^p+tB^q
$$
for all $t\in[0,1]$. Now one can choose the right $t\in[0,1]$ to get $(**)$. 
