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Let $(X,\rho)$ be a metric space. Suppose $f$ to be a real-valued funtion on $X$. A function $w: \left[0,\infty\right)\rightarrow \left[0,\infty\right)$ is said to be a modulus of continuity of the function $f$, if $$|f(x)-f(y)|\leqslant w(\rho(x,y))\;\;\; (x,y\in X).$$

My question is if the following sentence is true:

Assume that $K$ be a compact subest of X and let $f:K\rightarrow \mathbb{R}$ be a continuous function. Then $f$ has a modulus of continuity $w$ which is increasing, concave and $\lim_{t\to 0} w(t)=w(0)=0$.

I know that the function $w$ given by $$w(t)=\sup_{\rho(x,y)=t} |f(x)-f(y)|\;\;\; (t\geq 0),$$ is a modulus of continuity of such $f$ and $\lim_{t\to 0} w(t)=w(0)=0$, and then we may set $w_1(t)=\sup_{s\leq t} w(s)$ which is also increasing, but is possible to obtain the concavity too?

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As $w$ is only defined by $\geq$, you can make it anything you like, as long as it grows fast enough. – Karolis JuodelÄ— Sep 1 '12 at 17:17
You can take the concave envelope of $\omega$, I wrote a brief explanation here – user31373 Sep 1 '12 at 19:50
Thank you very much! – dawid Sep 2 '12 at 15:17

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