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If $f(x)$ be a function such that $f''(x)\geq 0$ on $(a, b)$ then applyint MVT show that for all $x,y\in (a, b)$ and for all $\alpha \in [0,1]$ $$f(\alpha x+(1-\alpha)y)\leq \alpha f(x)+(1-\alpha)f(y).$$

I know that I have to use Lagrange's MVT, but I don't understand how to use it here. Please help.

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    https://en.wikipedia.org/wiki/Jensen's_inequality – 2017-01-13
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    There is a simple picture which will help you! – 2017-01-13
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    Informally: $f''(x)$ gives that $f'(x)$ is increasing, hence the slope of the tangent lines increases when travelling from left to right. In particular, that implies the graph of $f(x)$ lies above the graph of any tangent, so $f(x)$ is a convex function. – 2017-01-13
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    Both conditions can be treated as independent definitions of a convex function. – 2017-01-13
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    @JackD'Aurizio Would you please elaborate in answer style – 2017-01-13
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    @StubbornAtom: I do not agree with that. Convexity does not imply that $f''$ exists, but $f''\geq 0$ implies convexity. So they are not equivalent conditions, and they are *very* dependent. – 2017-01-13
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    @JackD'Aurizio My bad, you are spot on. – 2017-01-13

2 Answers 2

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For simplicity, let $x

1

Fix $x,y\in[a,b]$ (let us suppose that $x

$$\phi(\alpha)=\alpha f(x)+(1-\alpha)f(y)-f(\alpha x+(1-\alpha)y)$$

Now compute :

$$\phi'(\alpha)=f(x)-f(y)-(x-y)f'(\alpha x+(1-\alpha)y)$$

By MVT, there exists some $c\in(x,y)$ such that $f(x)-f(y)=(x-y)f'(c)$;

Thus :

$$\phi'(\alpha)=(x-y)(f'(c)-f'(\alpha x+(1-\alpha)y))$$

Let us denote :$$\alpha_0=\frac{y-c}{y-x}$$

Since $f'$ is increasing, we can conclude that $\phi$ is increasing for $x\in[0,\alpha_0]$ and decreasing for $x\in[\alpha_0,1]$.

But $\phi(0)=\phi(1)=0$, so that $\forall\alpha\in[0,1],\,\phi(\alpha)\ge0$, which completes the proof.

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    Actually your $c$ is between $x$ and $y$. – 2017-01-13
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    Oops ! Thank you. I am editing it right now. – 2017-01-13