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Let $f\colon [a,b] \rightarrow \mathbb R$ be continuous and convex.

Let $m \colon [a,b] \rightarrow \mathbb R$ and $m(x) = \max \left\{f(y): y \in [a,x] \right\}$. Prove that $m$ is convex

I have no idea how to begin the proof. Do you have any hints?

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dyrAnd
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    Consider $s = \sup { c \in [a,b] : f(c) \leqslant f(a)}$ and the intervals $[a,s]$ and $[s,b]$. – Daniel Fischer Aug 14 '15 at 10:53
  • Inside $[s,b]$, $m(x) = f(x)$ and it is convex. On $[a,s]$, $m(x)$ is constant so it is also convex. When we take $x$ from $[a,s]$ and $y$ from $[s,b]$. I think that section between $x$ and $y$ is above graph of a function, yet I still do not know how to prove it. – dyrAnd Aug 14 '15 at 11:06
  • What can you say about the secants for $x,y$ and for $s,y$? – Daniel Fischer Aug 14 '15 at 11:12
  • Perhaps better: $m(x) = \max { f(x), f(a)}$. Show the maximum of two convex functions is convex. – Daniel Fischer Aug 14 '15 at 11:27

2 Answers2

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*What I had written first required differentiability I apologize. This is the same idea although quite unelegant in this form. Thanks Vim for pointing out my error.

First understand why the definition of a convex function means that its graph between any two points $[c,d]$ is below the line through those two points.

Observe that $f$ cannot decrease after it has been increasing. If $f$ is convex in an interval [c,d] if $max \{ f(x) \ | \ x\in [c,d]\}=f(x_m)$ for $c<x_m<d$ let $c<x_l<x_m<x_r<d$. Then the line through $x_l$ and $x_r$ lies below $f(x)$.

This means that if $f(x)$ is an increasing function in any interval $[a,u]$ it is an increasing function in the whole domain $[a,b]$.

CASE 1: If $f(x)$ is an increasing function $m(x)=f(x)$.

CASE 2: If $f(x)$ is a decreasing function in an interval $[a,u]$ we have two subcases.

2.1 If$f(a)=max\{ f(x) \ | \ x\in[a,b]\}$, then $m(x)=f(a)=const$. Hence it is convex.

2.2 There is a point $x_a\in [a,b)$ such that f(x_a)=f(a).

In that case $$m(x)=\cases{f(a)=const \ \ \ x\in [a,x_a] \\ f(x) \ \ \ \ \ x\in (x_a, b]}$$

Now we must prove that $m(x)\leq m(c)+ (m(d)-m(c))\frac{x-c}{d-c}$, for $x\in [a,b]$ for all $c,d\in [a,b]$.

We already have the cases where $c,d \in [a,x_a]$ or $a,b \in[x_a,b]$.

So we have to prove it for $c\in [a,x_a)$ and $d\in(x_a,b]$ which is readily obvious.

$$\forall x\in[a,x_a] \ \ m(x)=m(a)\leq m(c)+ (m(d)-m(c))\frac{x-c}{d-c}$$ $$\forall x\in(x_a,b] \ \ m(x) = f(x) \leq m(c)+ (m(d)-m(c))\frac{x-x_a}{d-x_a}\leq m(c)+ (m(d)-m(c))\frac{x-c}{d-c}$$.

Since in that interval $$\frac{x-x_a}{d-x_a}\leq \frac{x-c}{d-c}$$

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  • Indeed I do not know that even f is, I saw the more convenient explanation with the second derivative and went for it. I will edit. – Mr.P Aug 14 '15 at 11:48
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Fix $a \leq x \leq b$. By Theorem 1, Section 2, $f$ attains its maximum on $[a,x]$ either at $a$ or at $x$. As a consequence, $$ m(x)=\max\{f(a),f(x)\}. $$ Now observe that a constant function is always convex, and apply this result.

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