# Proving analytical statement, Intermediate Value Theorem

Let's define $f$ as a continuous function with $f:[0;2] \to \mathbb{R}$ and $f(0) = f(2)$.

Now, I want to show that:

$$\exists x_0 \in [0;1]:f(x_0) = f(x_0 + 1)$$ I tried to plot a few functions in order to construct a counterexample, but it seems that this statement really is true.

Unfortunately, I don't think I'm entirely sure why this works, yet. My current guess is that it has something to do with the Intermediate Value Theorem, as $f$ is continuous. In other words: If our function value 'goes up', it'll have to 'come back down' eventually (and vice versa), since it still needs to fulfill $f(0)=f(2)$.

Can someone help me prove this statement?

• Yes, you should set this up as an application of the Intermediate Value Theorem. How do you usually use that theorem to prove a certain equation has a solution? Commented May 14, 2014 at 21:03
• Here is a related post. Commented May 14, 2014 at 21:07
• @Ted Well, I'm new to this theorem and haven't seen it applied a lot. But from what my professor told me, it would be a smart idea to consider the function $g:[0;1] \to \mathbb{R}, g(x) = f(x+1)-f(x)$. Commented May 14, 2014 at 21:09
• Precisely, @chiru. If you want to show there's a place on $[0,1]$ where $\cos x=x$, you consider $g(x)=\cos x - x$ and you apply the Intermediate Value Theorem, after you argue that $g$ is continuous. Commented May 14, 2014 at 21:11
• @chiru Smart ideas are worth persuiing. What is $g(0)$? What is $g(1)$? You don't know eithre of these values, but you know $g(0)+g(1)$, don't you? Commented May 14, 2014 at 21:15

Consider the following function on $[0,1]$,

$$g(x) = f(x+1) - f(x)$$

where we assume $f(0) = f(2)$. We'd like to show there is some $x_0 \in [0,1]$ such that $g(x_0)=0$. So if we can show $g(x) >0$ on some interval and $g(x) <0$ somewhere as well, since $g(x)$ is continuous it must pass $0$. This is the context of the intermediate value theorem. So let's consider $x=0$ and $1$, we see that

$$g(0) = f(1) -f(0) \quad \& \quad g(1) = f(2) -f(1) = f(0) - f(1) = -g(0)$$

Thus we see that $g(x)$ does indeed change sign. So by the intermediate value theorem we know that it must be zero somewhere.

• Thank you! I understand this proof, but it misses the fact that we can easily assume that $g$ is continuous. We make this assumption since continuity won't suddenly disappear when adding, subtracting or even multiplying a function with another, which is basically how $g(x)$ has been defined. Thanks a lot! :) Commented May 14, 2014 at 21:41
• I mentioned that $g$ is continuous in there for IVT, but we know this since $f$ is continuous (and addition of continuous functions is continuous)
– Jeb
Commented May 14, 2014 at 21:46

Define $$g(x)=f(x)-f(x+1)$$ for $x$ in $[0,1$] then you have that in 0 is positive but in 1 is negative and then by the MVT is done.

• Can it happen that in 0 and 1 both vanishes. And the problem is done too. Commented May 14, 2014 at 21:19

$f$ is a continuous function $f\colon [0;n] \to \mathbb{R}$, $f(0) = f(n)$ then there are $(x_i,y_i)(i=1,2,\dotsc,n)$ such that $y_i-x_i$ is positive integer,and $f(x_i) = f(y_i)$.

One way is mathematical induction

$g(x) = f(x+1) - f(x)$

there is $\epsilon$ such that $g(\epsilon) = 0$. Consider

$$$$h(x)=\begin{cases}f(x)& 0\leq x\leq\epsilon\\f(x+1)&\epsilon\lt x\leq n\end{cases}$$$$

now $h(0)=h(n)$