Continuous extension of $f(x)=x\log(x)$: Show $\lvert f(a)-f(b)\rvert\leq\lvert f(b-a)\rvert$ for $0

Question

I am considering the function $$ f(x)=\begin{cases}x\log x, & x>0\\ 0, & x=0\end{cases} $$ on an interval $[0,\varepsilon]$ for small $\varepsilon>0$.

I would like to argue that for $x,y\in [0,\varepsilon]$ with $0<a<b$ one has $$ \lvert f(a)-f(b)\rvert\leq\lvert f(b-a)\rvert. $$

I think, in order to show this, one could use that near $x=0$ the function is convex and $f(0)=0$.

This immediately implies superadditivity, that is, $$ f(a+b)\geq f(a)+f(b). $$

Can one use this to prove the desired statement?

I do not see it.

I see! Then my idea is wrong. Is there another way to see the statement (or is it false)? — John_Doe, Feb 15 '21 at 14:27
My previous remark was wrong. I have taken the liberty to restore your original question. — Martin R, Feb 15 '21 at 14:41

Martin R · Accepted Answer · 2021-02-15T14:44:35.163

1

Your idea works. $f$ is convex with $f(0) = 0$ and therefore superadditive. $f$ is decreasing on $[0, 1/e]$, so that for $0 \le a < b \le 1/e$ $$ |f(a) - f(b)| = f(a) - f(b) = f(a) - f(a+b-a) \\ \le f(a) - (f(a) + f(b-a)) = -f(b-a) = |f(b-a)|\, . $$

edited Feb 15 '21 at 14:44

answered Feb 15 '21 at 14:40

Martin R

113,040

What is $1/e$? Is this my $\varepsilon$? – John_Doe Feb 15 '21 at 14:41
@John_Doe: $e$ is the Euler number. For $x < 1/e$ is $f'(x) = \log(x) + 1 < 0$ so that $f$ is decreasing (assuming that you mean the natural logarithm). – Martin R Feb 15 '21 at 14:43
I do see your point. – John_Doe Feb 15 '21 at 14:49

Continuous extension of $f(x)=x\log(x)$: Show $\lvert f(a)-f(b)\rvert\leq\lvert f(b-a)\rvert$ for $0

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