Maximum value at an endpoint

Question

Let $f: R \to R$ be a continuous function that satisfies $\forall x \in R$ $$f(x) \leq \frac{f(x - h) + f(x + h)}{2}$$

$\forall h > 0$. Show that the maximum value of f on any bounded closed interval $[a,b]$ is attained at one of the endpoints. That is, either $f(a)$ or $f(b)$ is the maximum value of f on the interval [a,b].

I'm not actually convinced the maximum is at an endpoint. I'm honestly not sure how to analyze this function. Can anyone give me some insight here? Thank you!

Note: This is an an analysis class where we have not covered differentiation. So we cannot use that.

Answer 1

If the maximum is not attained at the endpoints, then it is attained at some $c \in (a,b)$, with $f(a) < f(c)$ and $f(b) < f(c)$. Assume without loss of generality that $c$ is no closer to $b$ than it is to $a$, and let $h = c-a$. Then $f(c-h) = f(a) < f(c)$, and since $b$ is no closer to $c$ than it is to $a$, we have $c+h \in [a,b]$, which implies that $f(c+h) \leq f(c)$. Putting these together,

$$f(c) \leq \frac{f(c-h)+f(c+h)}2 < \frac{f(c)+f(c)}2 = f(c),$$

a contradiction.

Answer 2

Suppose the maximum value $M$ of $f$on $[a,b]$ is attained at a point $c$. If $c=1$ or $c=b$ we are done. Suppose $a<c<b$. Let $h$ be the minimum of the numbers $c-a$ and $b-c$. Then $M=f(c)\leq \frac {f(c-h)+f(c+h)} 2\leq \frac {M+M} 2=M$. This forces $f(c-h)=f(c+h)=M$ I will leave it to you to check that $\{c-h,c+h\}$ contains one of the end points $a,b$.