How to prove $f(x)=\log\left(\displaystyle\sum_{i=1}^n e^{x_i}\right)$ is a convex function?
EDIT1: for above function $f(x)$, following inequalities hold:
$$\max\{x_1,x_2,\ldots,x_n\}\leqslant f(x)\leqslant\max\{x_1,x_2,\ldots,x_n\}+\log n$$
and I have tried proving its convexity via definition of a convex function with above inequalities, but that didn't work.
EDIT2: I have posted my answers below.
Best Answer
Proof:
Let $u_i=e^ {x_i} ,v_i=e^ {y_i}$. So $f(\theta x+(1-\theta)y)=log(\sum_ {i=1}^n e^{\theta x_i + (1-\theta)y_i})=log(\sum_ {i=1}^n u_i^ \theta v_i^{(1-\theta)})$
From Hölder's inequality:
$$\sum_ {i=1}^n x_iy_i \le (\sum_ {i=1}^n|x_i|^p)^{\frac{1}{p}} \cdot (\sum_ {i=1}^n|x_i|^q)^{\frac{1}{q}}$$ where $1/p+1/q=1$.
Applying this inequality to $f(\theta x+(1-\theta)y)$: $$log(\sum_ {i=1}^n u_i^ \theta v_i^{(1-\theta)}) \le log[(\sum_ {i=1}^n u_i^ {\theta \cdot \frac{1}{\theta}})^ \theta \cdot (\sum_ {i=1}^n v_i^ {1-\theta \cdot \frac{1}{1-\theta}})^ {1-\theta}]$$ Right formula can be reduced to:
$$\theta log\sum_ {i=1}^n u_i+(1-\theta)log\sum_ {i=1}^n v_i$$
Here I regard $\theta$ as $\frac{1}{p}$ and $1-\theta$ as $\frac{1}{q}$.
So I achieve that $f(\theta x+(1-\theta)y) \le \theta f(x) + (1-\theta)f(y)$.