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I want to prove that if $f:\mathbb{R}\to\mathbb{R}^{+}\cup\{0\}$ is a strictly convex function, that is, for every $\alpha\in(0,1)$ $$ f(\alpha x+(1-\alpha)y)< \alpha f(x)+(1-\alpha)f(y) $$

if $f(0)=0$ and if $d$ is a metric over $X$, then $f\circ d$ is also a metric over $X$. I know $x=y$ $\Leftrightarrow$ $(f\circ d)(x,y)=0$. I just need to prove that $$ f(d(x,y))\leq f(d(x,z))+f(d(y,z) ) $$ for every $x,y,z\in X$, but i don't quite get it.

If I'm correct, because $f$ is non-negative then it must be increasing in $[0,\infty)$, and so $$ f(d(x,y))\leq f(d(x,z)+d(y,z)) $$ but that's not the desired inequality. On the other hand, taking $\alpha=\frac{1}{2}$, $$ 2f\left(\frac{1}{2}d(x,y)\right)\leq2f\left(\frac{1}{2} d(x,z)+\frac{1}{2}d(y,z)\right)< f(d(x,z))+f(d(y,z)) $$

I haven't got further than that.

Glutch_Dromedarius
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1 Answers1

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This is not true. For example take the metric space on three points $x_1$, $x_2$, and $x_3$ with $d(x_1, x_2) = 1$, $d(x_2, x_3) = 1$, and $d(x_2, x_3) = 2$ and the strictly convex function $f(x) = e^x$. Then $e^{2} = e^{d(x_1, x_3)} > e^{d(x_1, x_2)} + e^{d(x_2, x_3)} = 2 e$.

This is true when we replace the hypothesis of strict convexity with concavity: for a proof of this case see this answer: Prove that this is a metric space

Keeley Hoek
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