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Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f(x,y)=\ln(e^{5x}+e^{2y})$ for all $(x,y)\in \mathbb{R}^2$.

According to what I've seen (and taught by my teachers), a function $f$ is convex if it satisfies

$(\forall x,y\in \mathbb{R}^2, 0\leq t\leq 1)$

$f(t\cdot x+(1-t)\cdot y)\leq t\cdot f(x) +(1-t)\cdot f(y)$

I'm trying to prove the given function is convex by simply choosing $a=(x_1,y_1),b=(x_2,y_2)\in \mathbb{R}^2$ and applying to the lefthand side of the inequality, yet after playing around with the equation, I cannot seem to find a way to prove the required inequality.

Any ideas/path of thoughts would be of assistance.

amWhy
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Aishgadol
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2 Answers2

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In order to be a convex function, a sufficient condition for a $f \in C^2(\mathbb R^2)$ is that the Hessian matrix of $f$ be positive semidefinite for every $(x,y)\in \mathbb R^2$. Now $$ H(x,y) = \left( {\begin{array}{*{20}c} {\frac{{25e^{5x + 2y} }}{{\left( {e^{5x} + e^{2y} } \right)^2 }}} & { - \frac{{10e^{5x + 2y} }}{{\left( {e^{5x} + e^{2y} } \right)^2 }}} \\ { - \frac{{10e^{5x + 2y} }}{{\left( {e^{5x} + e^{2y} } \right)^2 }}} & {\frac{{4e^{5x + 2y} }}{{\left( {e^{5x} + e^{2y} } \right)^2 }}} \\ \end{array}} \right) $$ and its eigenvalues are $$ \begin{array}{l} \lambda _1 = \frac{{29e^{5x + 2y} }}{{\left( {e^{5x} + e^{2y} } \right)^2 }} > 0\,\,\,\,\forall \left( {x,y} \right) \in \mathbb R^2 \\ \lambda _2 = 0 \\ \end{array} $$ This is how to say that $H(x,y)$ is positive semidefinite, whence the thesis.

Luca Goldoni Ph.D.
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Another method with less computations, but only for this type of function.

Observe that $f(x,y) = 5x + \ln(1+e^{2y-5x})$. Since the functions $(x,y) \mapsto 5x$ and $(x,y) \mapsto 2y-5x$ are affine (and even linear), the convexity of $f$ follows from the convexity of the function $g : z \mapsto \ln(1+e^z)$, which follows from the fact that $$g'(z) = \frac{e^z}{1+e^z} = \frac{1}{e^{-z}+1}$$ is an increasing function of $z$.

Christophe Leuridan
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