Why is log-of-sum-of-exponentials $f(x)=\log\left(\sum_{i=1}^n e^ {x_i}\right)$ a convex function for $x \in\mathbb R^n$? 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.
 A: Another way to prove the convexity of this function is to use the Jensen's Inequality which states that a function is convex if and only if 
$$f(tX+(1-t)Y) \le t f(X) + (1-t)f(Y)$$
Now let $X$ be represented by the vector $({X_1, X_2, X_3,... X_n})$,
and let $Y$ be represented by the vector $({Y_1, Y_2, Y_3,... Y_n})$
Let $t = \dfrac{1}{2}$
$$f(tX+(1-t)Y) = \log\left(\sum_{i=1}^{n} e^{\frac{X_i+Y_i}{2}}\right)$$
$$\text{RHS} = \frac{1}{2} \log\left(\sum_{i = 1}^{n} e^{X_i}\right)+ \frac{1}{2} \log\left(\sum_{i = 1}^{n} e^{Y_i}\right)$$
$$\text{RHS} = \frac{1}{2} \log\left(\sum_{i = 1}^{n} e^{X_i}\sum_{i = 1}^{n} e^{Y_i}\right)$$
RHS contains more cross product terms than the LHS thus making it larger than LHS and hence the function is convex.
A: This answer is similar to the answer written by @Nicholas, but I'm including more details.
A nice fact about the logSumExp function $f$ is that its gradient is the softmax function $S$:
$$
\nabla f(x) = S(x) = \begin{bmatrix} \frac{e^{x_1}}{e^{x_1} + \cdots + e^{x_n}} \\ \vdots \\ \frac{e^{x_n}}{e^{x_1} + \cdots +e^{x_n}} \end{bmatrix}.
$$
The Hessian of $f$ is the matrix $S'(x)$, and a nice fact about the softmax function is that
$$
S'(x) = \text{diag}(S(x)) - S(x) S(x)^T.
$$
If we can show that $S'(x)$ is positive semidefinite, it will follow that $f$ is convex.
Edit: At this point, I recommend reading @Bruno-84’s proof, which is superior to the argument that I gave below.
Original argument:
In other words, we need to show that if $v \in \mathbb R^n$, then $v^T S'(x) v \geq 0$. But notice that
\begin{align}
& v^T S'(x) v \geq 0 \\
\iff & v^T \text{diag}(S(x)) v \geq v^T S(x) S(x)^T v \\
\iff & \sum_{i=1}^n \left( \frac{e^{x_i}}{e^{x_1} + \cdots + e^{x_n}}\right) v_i^2 \geq (S(x)^T v )^2 \\
\iff & \sum_{i=1}^n \left( \frac{e^{x_i}}{e^{x_1} + \cdots + e^{x_n}}\right) v_i^2 \geq \left( \sum_{i=1}^n v_i \cdot \frac{e^{x_i}}{e^{x_1} + \cdots + e^{x_n}} \right)^2 \\
\iff & \left(\sum_{i=1}^n e^{x_i} v_i^2 \right) \left(\sum_{i=1}^n e^{x_i}\right) \geq \left(\sum_{i=1}^n v_i e^{x_i} \right)^2
\end{align}
This last inequality is true, as can be seen by applying the Cauchy-Schwarz inequality to the vectors
$$
a = \begin{bmatrix} \sqrt{e^{x_1}} \\ \vdots \\ \sqrt{e^{x_n}} \end{bmatrix},
\quad b = \begin{bmatrix} v_1 \sqrt{e^{x_1}} \\ \vdots \\ v_n \sqrt{e^{x_n}} \end{bmatrix}.
$$
A: 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:
$$\sum_ {i=1}^n x_iy_i \le (\sum_ {i=1}^n|x_i|^p)^{\frac{1}{p}} \cdot (\sum_ {i=1}^n|y_i|^q)^{\frac{1}{q}}$$ where $1/p+1/q=1$.
Applying this inequality to $f(\theta x+(1-\theta)y)$:
$$log(\sum_ {i=1}^n u_i^ \theta v_i^{(1-\theta)}) \le 
log[(\sum_ {i=1}^n u_i^ {\theta \cdot \frac{1}{\theta}})^ \theta \cdot 
(\sum_ {i=1}^n v_i^ {1-\theta \cdot \frac{1}{1-\theta}})^ {1-\theta}]$$
The 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 $f(\theta x+(1-\theta)y) \le \theta f(x) + (1-\theta)f(y)$.
A: For a multivariate function to be convex, it's equivalent to show that its Hessian matrix is positive semi-definite. That is, you can calculate $\nabla^2 f(\mathbf{x})$ here and show it is positive semi-definite.
This can be proved using Cauchy Schwarz inequality as shown here.
A: I have a preference for proving that the Hessian matrix is non negative as fully developed by @littleO.
Once you have shown that the gradient is the softmax function $S(x)$ and the Hessian matrix has the expression
$$
\nabla^2 f(x) = \operatorname{diag}(S(x)) - S(x)S(x)^T,
$$
you can simply use the convexity of the real map $t\mapsto t^2$ to conclude that
$\nabla^2 f(x)$ is non negative: Since $S(x)^T v$ is a convex linear combination of the coordinates of the vector $v$ (that is $S(x)_k\geq 0$ and $\sum_{k=1}^n S(x)_k =1$), one has
$$
v^T S(x) S(x)^T v 
= (S(x)^T v)^2
= \left(\sum_{k=1}^n S(x)_k v_k\right)^2
\leq \sum_{k=1}^n S(x)_k v_k^2 = v^T \operatorname{diag}(S(x)) v.
$$
A: It is enough to show that 
$$\frac{1}{2} \log (\sum \exp x_i) + \frac{1}{2}\log (\sum \exp y_i)\ge \log (\sum \exp\frac{x_i+y_i}{2})$$
or, with the substitution $\exp\frac{x_i}{2} = a_i$, $\exp\frac{y_i}{2} = b_i$
$$(\sum a_i^2)^{\frac{1}{2}}(\sum b_i^2)^{\frac{1}{2}}\ge \sum a_i b_i$$
