Show that $\log(e^{x_1} + e^{x_2})$ is convex.

Question

Do I need to find Hessian matrix? or is there any other method ?

Answer 1

The function $\;f(x,y)=\log(e^x+e^y)\;$ is convex iff the function

$$\;g(t):=f(\vec x+t\vec y)=\log\left(e^{x_1+ty_1}+e^{x_2+ty_2}\right)\;$$ is convex, but:

$$\begin{align*}g'(t)&=\frac{y_1e^{x_1+ty_1}+y_2e^{x_2+ty_2}}{e^{x_1+ty_1}+e^{x_2+ty_2}}\\{}\\g''(t)&=\frac{\left(y_1^2e^{x_1+ty_1}+y_2^2e^{x_2+ty_2}\right)(e^{x_1+ty_1}+e^{x_2+ty_2})-(y_1e^{x_1+ty_1}+y_2e^{x_2+ty_2})^2}{(e^{x_1+ty_1}+e^{x_2+ty_2})^2}\end{align*}\;$$

Answer 2

Let $f(x_1,x_2)=\log\big(e^{x_1}+e^{x_2} \big)$. No you do not necessarily need to calculate the Hessian. You can prove that for every curve $\alpha(t)=(x_1(t),x_2(t))$ such that $\alpha(0)=(x_1,x_2)$ for all $(x_1,x_2)$ and $t$ $$ g^{\prime\prime}(t)=(f\circ \alpha)^{\prime\prime}(t)\geq 0 $$ But calculating the Hessian of this function is not as laborious as well.We need prove that $\quad \forall v=(v_1,v_2)\in\mathbb{R}^2$ \begin{align} v \cdot \mbox{Hess}f(x_1,x_2)\cdot v^T = & \left[v_1,v_2 \right]\cdot \left[ \begin{array}{cc} \frac{\partial\, f(x_1,x_2)}{\partial x_1\partial x_1} & \frac{\partial\, f(x_1,x_2)}{\partial x_2\partial x_1} \\ \frac{\partial\, f(x_1,x_2)}{\partial x_1\partial x_2} & \frac{\partial\, f(x_1,x_2)}{\partial x_2\partial x_2} \end{array} \right] \cdot \left[\begin{array}{c}v_1\\v_2 \end{array}\right]\geq 0, \\ =&\frac{\partial\, f(x_1,x_2)}{\partial x_1\partial x_1}v_1v_1 + \frac{\partial\, f(x_1,x_2)}{\partial x_1\partial x_2}v_1v_2 \\ +&\frac{\partial\, f(x_1,x_2)}{\partial x_2\partial x_1}v_2v_1 + \frac{\partial\, f(x_1,x_2)}{\partial x_2\partial x_2}v_2v_2 \geq0 \end{align} We have
\begin{align} v \cdot \mbox{Hess}f(x_1,x_2)\cdot v^T = & \left[v_1,v_2 \right]\cdot \left[ \begin{array}{cc} \frac{e^{x_1}\cdot e^{x_2}}{e^{x_1}+e^{x_2}} & -\frac{e^{x_1}\cdot e^{x_2}}{e^{x_1}+e^{x_2}} \\ -\frac{e^{x_1}\cdot e^{x_2}}{e^{x_1}+e^{x_2}} & \frac{e^{x_1}\cdot e^{x_2}}{e^{x_1}+e^{x_2}} \end{array} \right] \cdot \left[\begin{array}{c}v_1\\v_2 \end{array}\right] \\ =&\frac{e^{x_1}\cdot e^{x_2}}{e^{x_1}+e^{x_2}}\left(v_1^2-2v_1v_2+v_2^2 \right) \\ =& \frac{e^{x_1}\cdot e^{x_2}}{e^{x_1}+e^{x_2}}\left(v_1-v_2 \right)^2\geq 0 \end{align}

Partial derivatives \begin{align} \frac{\partial\, f(x_1,x_2)}{\partial x_1\partial x_1} = & \frac{\partial\,}{\partial x_1\partial x_1}\log( e^{x_1}+e^{x_1}) = \frac{\partial\,}{\partial x_1}\left(\frac{e^{x_1}}{e^{x_1}+e^{x_2}}\right) \\ = & \frac{(e^{x_1})_{x_1}(e^{x_1}+e^{x_2})-(e^{x_1})(e^{x_1}+e^{x_2})_{x_1}}{(e^{x_1}+e^{x_2})^2} = \frac{e^{x_1}e^{x_2}}{(e^{x_1}+e^{x_2})^2} \end{align} analogously $$ \frac{\partial\, f(x_1,x_2)}{\partial x_2\partial x_2}=\frac{e^{x_1}e^{x_2}}{(e^{x_1}+e^{x_2})^2} $$ and $$ \frac{\partial\, f(x_1,x_2)}{\partial x_2\partial x_1}= \frac{\partial\, f(x_1,x_2)}{\partial x_1\partial x_2} = \frac{\partial\,}{\partial x_2\partial x_1}\log( e^{x_1}+e^{x_1}) = \frac{\partial\,}{\partial x_2}\left(\frac{e^{x_1}}{e^{x_1}+e^{x_2}}\right) = -\frac{e^{x_1}\cdot e^{x_2}}{e^{x_1}+e^{x_2}} $$