Let $f:\mathbb R\to \mathbb R$ be such that for all $x,y \in\mathbb R,\ \ f(y) \geq f (x) + f'(x)(y−x).$
How to prove that $f$ is a convex function?
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Hint: By hypotesis $$f(y)\geqslant f(x)+f'(x)(x-y),\qquad f(x)\geqslant f(y)+f'(y)(y-x)$$use this to prove that $f'(x)-f'(y)\geqslant 0$. Take then $z=\lambda x+(1-\lambda)y$ (with $\lambda\in(0,1)$) and apply the hypotesis, again to get $$f(z)\geqslant f(x)+f'(x)(z-x),\qquad f(z)\geqslant f(y)+f'(y)(z-y)$$ once here, multiply the first of the last two inequalities by $\lambda$ and the second by $1-\lambda$ and sum the two, then you get the conclusion (with a bit of manipulating) – Alessandro Apr 11 '21 at 15:44
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Thanks @Alessandro! Can you please elaborate a bit what do you mean by "and sum the two, then you get the conclusion (with a bit of manipulating)" – user6001037 Apr 11 '21 at 15:49
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Well, if you have two inequalities $a\geqslant b$ and $c\geqslant d$, you can sum them in $a+c\geqslant b+d$ – Alessandro Apr 11 '21 at 15:50
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Let $z = \alpha x + (1-\alpha) y$ where $0 \le \alpha \le 1 $. By hyopthesis we have: \begin{align} &f(x) \ge f'(z)(x-z) + f(z) = f'(z)\bigg[x - \alpha x + (1-\alpha) y\bigg] + f(z) = (1-\alpha)f'(z)(x- y) + f(z)\\ \implies & \alpha f(x) \ge \alpha(1-\alpha)f'(z)(x- y) + \alpha f(z)\qquad\text{(1)}. \end{align} Again by hypothesis: \begin{align} &f(y) \ge f'(z)(y-z) + f(z) = f'(z)\bigg[y - \alpha x + (1-\alpha) y\bigg] + f(z) = -\alpha f'(z)(x- y) + f(z)\\ \implies & (1-\alpha)f(y) = -\alpha(1-\alpha) f'(z)(x- y) + (1-\alpha)f(z)\qquad\text{(2)}. \end{align} Adding (1) and (2): \begin{align} \alpha f(x) + (1-\alpha)f(y) \ge \alpha f(z) + (1-\alpha) f(z) = f(z) = f(\alpha x + (1-\alpha)y). \end{align}
