Convex Function Inequality

Question

Let $f: I \to \mathbb{R}$ where $I$ is an interval. We say that $f$ is convex if for every $a,b\in I$ and every $\lambda : 0<\lambda < 1.\\$ Prove that for any $x\in (a,b)$ $f(\lambda b + (1-\lambda)a) \leq \lambda f(b) + (1-\lambda )f(a) \implies f(x) \leq \frac{(x-a)}{(b-a)}f(b) + \frac{(b-x)}{(b-a)}f(a)\\$

I can see how this works clearly geometrically but I am unsure where to start. It is clear that $\frac{x-a}{b-a}$ and $\frac{b-x}{b-a} \leq 1$. And by the triangle inequality that $f(x)\leq f(a) + f(b)$ but I don't know where to even begin. Can anyone help?

score 4 · Accepted Answer · answered Oct 21 '13 at 00:12

4

If $$\lambda= \frac{x-a}{b-a},$$ then $$1-\lambda= \frac{b-a}{b-a}-\frac{x-a}{b-a}=\frac{b-x}{b-a}.$$

Mathematical statements are usually Extremely general... the above inequality holds for $\textit{all}$ $0<\lambda<1,$ in particular even for $\lambda =\frac{x-a}{b-a},$ or $\lambda=\frac{b-x}{b-a}.$

answered Oct 21 '13 at 00:12

Maxim Gilula

1,485

Thanks a lot! This was all I needed to get it done – Kyle H. Oct 21 '13 at 01:39
Quick question though if you are still around...Does $\lambda <0, \lambda >1$ change the direction of the inequality? – Kyle H. Oct 21 '13 at 02:02
Try to draw $\lambda f(a)+(1-\lambda)f(b)$ as a function of $\lambda$ for $f$ convex. Try for example $f(x)=x^2.$ – Maxim Gilula Oct 21 '13 at 04:12

Convex Function Inequality

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