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Let ‎$‎f:[a, b]‎‎\rightarrow ‎‎\mathbb{R}‎‎‎‎$ ‎be a ‎function. My ‎q‎uestion is:‎

If ‎$‎‎‎f‎$‎‎‎‎‎‎‎‎‎ ‎is a ‎convex function‎‎‎‎ ‎and ‎‎$‎f(x)\geq f(a)‎$ ‎for ‎all ‎‎$‎a\leq x\leq b‎$‎, then ‎$‎f‎$ ‎is ‎increasing‎‎‎‎‎‎‎‎‎‎.‎‎

‎ I ‎know ‎that ‎if ‎‎$‎f‎$ ‎is ‎differentiable ‎on ‎‎$‎(a, b)‎$‎, then ‎$‎f‎$ ‎is ‎convex ‎if ‎and ‎only ‎if ‎‎$‎f^\prime$ ‎is ‎increasing. ‎‎B‎ut ‎here ‎we ‎do not ‎have ‎the differentiability.

‎Anyone ‎can ‎help ‎me. Thanks‎

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Theorem. Let $f$ be a convex function in some interval $I$, and let $a<b<c$ be points in $I$. Then $$\frac{f(c)-f(a)}{c-a}\leq\frac{f(b)-f(a)}{b-a}\leq\frac{f(c)-f(b)}{c-b}$$

If $a<x_1<x_2\leq b$,then $$\frac{f(x_2)-f(x_1)}{x_2-x_1}\geq\frac{f(x_1)-f(a)}{x_1-a}\geq 0.$$ Moreover $f(x)\geq f(a)$ for $a\leq x\leq b$, so we have $f(x_1)\leq f(x_2)$ for any $a\leq x_1\leq x_2\leq b$, which completes the proof.

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Let $a\leq u\leq w\leq b$.

We can find $0 \leq \theta \leq 1$ such that $u = \theta a + (1 - \theta) w$. By convexity, $f(u) \leq \theta f(a) + (1-\theta) f(w)$. Since $f(a) \leq f(w)$, it follows that $f(u) \leq \theta f(w) + (1-\theta) f(w) = f(w)$.

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