Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous. Prove that there exists a point $c\in[a,b]$ such that: $f(c)\leq\frac{f(a)+f(b)}{2}$

Question

For this question, I figured out for the case when $f(a)=-f(b)$
when $f(a)=0$ we can take $c=a$ and if not we can use the intermediate value theorem. hi But for the case $f(a)\neq-f(b)$ we will get $|f(a)+f(b)|>0$. So I guess we can consider this as the $\epsilon$ for continuity. But couldn't proceed with

Answer 1

The image of $f([a,b])$ is an interval since $f$ is continuous, and ${{f(a)+f(b)}\over 2}$ is in the $[f(a),f(b)]$ or $[f(b),f(a)]$ which is contained in $f([a,b])$.

Answer 2

Note that $\min(x,y) \le {1 \over 2 } (x+y) \le \max(x,y)$.

Assume $x=f(a) \le y = f(b)$, then we obtain $f(a) \le {1 \over 2 } (f(a)+f(b)) \le f(b)$ and hence the intermediate value theorem shows that there is a $c$ such that $f(c) = {1 \over 2 } (f(a)+f(b))$.

Answer 3

You mention the IVT theorem in your post. Are you allowed to use that?

$\min(f(a),f(b)) = \frac {\min(f(a),f(b)) + \min(f(a),f(b))}2 \le \frac {f(a) + f(b)}2 \le \frac {\max(f(a),f(b)) + \max(f(a),f(b))}2 = \max(f(a), f(b))$.

So $\frac {f(a) + f(b)}2 $ is between $f(a)$ and $f(b)$ and by IVT there is a $c: a\le c \le b$ so that $f(c) = \frac {f(a) + f(b)}2 $

Are you sure you aren't actually supposed to prove the IVT?

Assume $f(a) < f(b)$

Let $A = \{c \in [a,b]| f(c) < \frac {f(a)+f(b)}2\}$. $A$ is non-empty because $a\in A$. And $b \not \in A$ and for all $c \in A$, $c < b$ so $A$ is bounded above by $b$.

Let $d = \sup A$. (Real number have the least upper bound property).

What is $f(d)$?

Answer 4

The answers above look somewhat complicated. If $f(c) >\frac {f(a)+f(b)} 2$ for all $c$ then $f(a) >\frac {f(a)+f(b)} 2$ and $f(b) >\frac {f(a)+f(b)} 2$ and this leads to the contradiction $\frac {f(a)+f(b)} 2 >\frac {f(a)+f(b)} 2$. Continuity is not required!