If ‎$‎f‎$ ‎is ‎a‎ convex function on ‎$‎[a, b]‎$‎, then ‎$‎f‎$ ‎is ‎increasing ‎on ‎$‎[a, b]‎$‎.‎

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

Theorem. Let $$f$$ be a convex function in some interval $$I$$, and let $$a be points in $$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,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.
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 $$f(u) \leq \theta f(w) + (1-\theta) f(w) = f(w)$$.