Showing that $f:\Bbb{R}^n\to\Bbb{R},\; x\mapsto\log(\sum_{1\leq i\leq n}e^{x_i})$ is convex

Question

I want to show that $f:\Bbb{R}^n\to\Bbb{R},\; x=(x_1,\dots,x_n)\mapsto\log(\sum_{1\leq i\leq n}e^{x_i})$ is convex. This is my attempt:
Let $x,y\in\Bbb{R}^n,\lambda\in[0,1]$.
I need to show that $f(\lambda x + (1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y):$

$f(\lambda x + (1-\lambda)y)=f\left( \begin{array}{c} \lambda x_1+(1-\lambda)y_1 \\ \dots\\ \lambda x_n+(1-\lambda)y_n \end{array} \right)=\log(e^{\lambda x_1+(1-\lambda)y_1}+\dots+e^{\lambda x_n+(1-\lambda)y_n})$

$=\log(e^{\lambda x_1}e^{(1-\lambda)y_1}+\dots+e^{\lambda x_n}e^{(1-\lambda)y_n})$

$\leq\log((e^{\lambda x_1}+e^{\lambda x_2}+\dots+e^{\lambda x_n})(e^{(1-\lambda) y_1}+e^{(1-\lambda) y_2}+\dots+e^{(1-\lambda) y_n}))$

$=\log(e^{\lambda x_1}+e^{\lambda x_2}+\dots+e^{\lambda x_n})+\log(e^{(1-\lambda) y_1}+e^{(1-\lambda) y_2}+\dots+e^{(1-\lambda) y_n})$

$\leq\log((e^{x_1}+e^{x_2}+\dots+e^{x_n})^\lambda)+\log((e^{ y_1}+e^{ y_2}+\dots+e^{ y_n})^{1-\lambda})$

$=\lambda\log(e^{x_1}+e^{x_2}+\dots+e^{x_n})+(1-\lambda)\log(e^{ y_1}+e^{ y_2}+\dots+e^{ y_n})$

$=\lambda f(x)+(1-\lambda) f(y)$

$\implies f$ is convex

Please let me know if I can make any improvements on this or if something is wrong. Thanks

An alternative proof would be to use that the sum of log-convex functions is again log-convex — Martin R, Sep 15 '20 at 17:10
In any case you should justify both inequalities used in your proof. — Martin R, Sep 15 '20 at 17:11
@MartinR ok thanks, I've actually never heard of the term "log-convex" but I'll have a look at that link that you sent — user619755, Sep 15 '20 at 17:13
See also Why log-of-sum-of-exponentials $f(x)=\log\left(\sum_{i=1}^n e^ {x_i}\right)$ is a convex function for $x \in\mathbb R^n$. — Martin R, Sep 15 '20 at 17:13
The second inequality is wrong. Note that $A^\lambda + B^\lambda > (A+B)^\lambda$ if $0 < \lambda < 1$ and $A, B > 0$. — Martin R, Sep 15 '20 at 17:17

score 3 · Answer 1 · answered Sep 16 '20 at 05:01

Unfortunately, your proof does not work because the second inequality is wrong. Apparently you are using $$ e^{\lambda x_1}+\ldots+e^{\lambda x_n} \le (e^{x_1}+\ldots+e^{x_n})^\lambda \, . $$ But that estimate does not hold, as can be seen from $$ \begin{align} 1 &= \frac{e^{x_1}}{e^{x_1}+\ldots+e^{x_n}} + \ldots + \frac{e^{x_n}}{e^{x_1}+\ldots+e^{x_n}} \\ &< \left( \frac{e^{x_1}}{e^{x_1}+\ldots+e^{x_n}}\right)^\lambda + \ldots + \left( \frac{e^{x_n}}{e^{x_1}+\ldots+e^{x_n}}\right)^\lambda \\ &= \frac{e^{\lambda x_1}+\ldots+e^{\lambda x_n}}{(e^{x_1}+\ldots+e^{x_n})^\lambda} \, . \end{align} $$

For a correct proof, see for example Why log-of-sum-of-exponentials $f(x)=\log\left(\sum_{i=1}^n e^ {x_i}\right)$ is a convex function for $x \in\mathbb R^n$ (which is based on Hölder's inequality).

Showing that $f:\Bbb{R}^n\to\Bbb{R},\; x\mapsto\log(\sum_{1\leq i\leq n}e^{x_i})$ is convex

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