Integral of a concave function

[talolard]

Homework Statement

Let $$f:[0,2]\rightarrow[0,\infty)$$ be continuous and non negative. Assume thaqt for any $$x,y\in[0,2]$$ and $$0<\lambda<1 f(\lambda x+(1-\lambda)y)\geq\lambda f(x)+(1-\lambda)f(y)$$. Given that f(1)=1 prove

$$\int_{0}^{2}f(x)dx\geq1$$

The Attempt at a Solution

I've sat for hours. I have zero inspiration. I need a gentle shove in the right direction please.

 

[Dick]

Consider the function g(x)=x for x in [0,1] and g(x)=2-x for x in [1,2]. Can you show f(x)>=g(x)?

 

[talolard]

I think i got it.
Define $$g(x):\mathbb{R}\rightarrow\mathbb{R}$$ . Such that $$g(x)=\begin{cases}<br /> x & 0\leq x\leq1\\<br /> 2-x & 1<x<2\end{cases}$$.

Assume that for some $$x\in[0,2] g(x)>f(x)$$ . Then $$g(x)=x>f(x)=f(x1+\left(1-x\right)y)\geq xf(1)+\left(1-x\right)f(y)=g(x)$$ +Something positive (Not precise here with the domain).

Then $$g(x)\geq g(x)+\epsilon$$ a contradiction. Thus we have that for all $$x\in[0,2] f(x)\geq g(x)$$

Then $$\int_{0}^{2}f(x)\geq\int_{0}^{2}f(x)=\int_{0}^{1}x+\int_{1}^{2}2-x=\frac{1}{2}+\left|2x-\frac{x^{2}}{2}\right|_{1}^{2}=\frac{1}{2}+4-2-2+\frac{1}{2}=1$$ Q.E.D.

 

[Dick]

I can't really figure out what your contradiction is all about. Did you draw a graph of g(x)? You know f(0)>=0 and f(1)=1. If you connect (0,f(0)) and (1,f(1)) with a line then your condition says the graph of f is above or on that line, right? Do you see how that proves f(x)>=g(x) on [0,1]? Can you translate the picture into a proof?