There are two definitions of a convex function over an interval $I$

Question

(i) For $x_1 < x < x_2$

we have

$$ \frac{f(x)-f(x_1)}{(x-x_1)} \le \frac{f(x_2)-f(x)}{(x_2-x)} \tag{1} $$

and

(ii) For every $x_1$ and $x_2$ in $I$ and every $λ: 0 < \lambda < 1$

we have

$$ f(\lambda x_1+(1-\lambda)x_2)\le \lambda f(x_1)+(1-\lambda)f(x_2)\tag{2} $$

Show that $(1)\implies(2)$

I have tried cross-multiplying the denominators in $(1)$ but I'm completely stuck.

Answer 1

After the cross multiplication, we have $(x_2-x_1)f(x)\leq(x_2-x)f(x_1)+(x-x_1)f(x_2)$

Hint: any number $x\in [x_1,x_2]$ can be written in the form $\lambda x_1+ (1-\lambda) x_2$, with $\lambda\in (0,1)$. I'm leaving this unproven as an exercise!

After that, our inequality becomes :$(x_2-x_1)f(x)\leq \lambda(x_2-x_1)f(x_1)+(1-\lambda)(x_2-x_1)f(x_2)$. By cancelling the common factor $(x_2-x_1)>0$ from both sides, you have what you want.

Answer 2

Assume that $x_1 < x < x_2$, and let $L$ be the function whose graph is the line segment joining the points $(x_1,f(x_1))$ and $(x_2, f(x_2))$.

$\qquad \qquad \qquad \qquad $

Of course, the slope of the segment on the left is equal to the slope of the segment on the right: $$ \frac{L(x) - f(x_1)}{x - x_1} = \frac{f(x_2) - L(x)}{x_2 - x}. $$

If we were to replace $L(x)$ with a greater number, then the term on the left would increase and the term on the right would decrease.

$\qquad \qquad \qquad \qquad$

So, if it were true that $f(x) > L(x)$, then we would have $$ \frac{f(x) - f(x_1)}{x - x_1} > \frac{L(x) - f(x_1)}{x - x_1} = \frac{f(x_2) - L(x)}{x_2 - x} > \frac{f(x_2) - f(x)}{x_2 - x} . $$

But this contradicts condition (i). Thus, condition (i) implies that $f(x) \leq L(x)$.