# How can the following be false based on the information about continuous functions?

Let $f\colon\mathbb{R}\to [0,\infty)$ be a continuous function. Then, what is the rationale behind saying that the following are false statements:

1. There exists $x\in\mathbb{R}$ such that $f(x)=\int_{-1}^{1}f(t)dt$.

2. There exists $x\in\mathbb{R}$ such that $f(x)=\frac{f(0)+f(1)}{2}$.

I think the second statement should be true due to the intermediate value theorem but I am not certain about the first statement. But my solution manual says both statements are false. What should be the rationale behind this reasoning?

• The question is not so much whether (1) and (2) are true or false as whether they are implied by the premise. Where do the problem and solutions come from? Jan 24 '18 at 10:04

You are correct: statement (2) is always true by the intermediate value theorem, since $\frac{f(0)+f(1)}{2}$ is between $f(0)$ and $f(1)$.

Statement (1) is not true for all $f$. A simple counterexample is the constant function $f(x)=1$.

The first statement reminds of the Mean Value Theorem for Integrals, but… The Mean Value Theorem for Integrals says that if $f$ is continuous on $[a,b]$, then there exists $x\in[a,b]$ such that $$f(x)=\frac{1}{b-a}\int_a^b f(t)\,dt.$$ In this case, $a=-1$ and $b=1$, so we can guarantee that there exists $x\in[-1,1]$ such that $$f(x)=\frac{1}{2}\int_{-1}^1 f(t)\,dt.$$ This is very similar to the given equality, but without the $1/2$. And that suggests that statement (1) is probably false. After realizing that, it's not difficult to come up with a counterexample.

As for statement (2), it's actually true by the Intermediate Value Theorem, since $f$ is continuous.