8
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ 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? $\endgroup$
    – PJTraill
    Commented Jan 24, 2018 at 10:04

2 Answers 2

19
$\begingroup$

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 $f(x)=1$.

$\endgroup$
0
15
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .