The convexity of $\frac{|x|}{1+|x|}$

Question

Let $f(x)=\frac{|x|}{1+|x|}$, the domain of $f$ is $\mathbb{R}$. Clearly, $f$ is even, and is nondecreasing in $[0,\infty)$ and is nonincresing in $(-\infty,0)$. It seems that $f(x)$ is convex, but I don't know how to prove it by definition. I also tried to prove convexity using convexity preserving operations, but failed.

Thanks in advance.

Quick beginner guide for asking a well-received question + please avoid "no clue" questions — Anne Bauval, May 23 '23 at 12:47

score 2 · Accepted Answer · answered May 23 '23 at 12:24

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The function is $C^2$ for $x \ne 0$, so you can analyze convexity separately in $]-\infty,0[$ and $]0, +\infty[$ just by studying the sign of the second derivative.The function is not convex or concave in intervals that contain $x=0$.

answered May 23 '23 at 12:24

PierreCarre

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Indeed, a graph reveals the non convexity clearly: https://i.stack.imgur.com/3ruu5.png – Calvin Khor May 23 '23 at 12:31
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Indeed (same prior comment above), so I think the post did not show enough effort to deserve an answer. – Anne Bauval May 23 '23 at 12:46

The convexity of $\frac{|x|}{1+|x|}$

1 Answers1