# Fascinating result of the Intermediate Value Theorem

This is problem #1 from chapter 3.9 in Advanced Calculus: Theory and Practice, my introductory real analysis textbook at Western Michigan University:

Suppose that $$f$$ is continuous on $$\left[0, 2\right]$$ and $$f(0) = f(2)$$. Prove that there exist $$x_1$$, $$x_2 \in \left[0, 2\right]$$ such that $$x_2 – x_1 = 1$$ and $$f(x_1) = f(x_2)$$.

Informally, this says there are two $$x$$-values exactly $$1$$ unit apart whose $$f$$ values are equal. This result isn’t all that obvious, and I liked it so much because it’s a great example of the type of abstract, theoretical result you learn to prove in mathematical analysis.

Recall that the Intermediate Value Theorem states that if $$f$$ is a continuous function on an interval $$\left[a, b\right]$$ and $$f(a) \neq f(b)$$, then for every $$C$$ between $$f(a)$$ and $$f(b)$$, there exists $$c \in (a, b)$$ such that $$f(c) = C$$. Often the Intermediate Value Theorem is stated as a specific case, where $$f(a) < 0$$ and $$f(b) > 0$$, in which case there exists $$c \in (a, b)$$ such that $$f(c) = 0$$. This is the case that will be relevant here. Now the solution to the problem:

Proof: Let $$g(x) = f(x+1) ~- f(x)$$, defined on $$\left[0, 1\right]$$. The function $$g$$ is continuous, and $$g(0) = f(1) ~- f(0)$$ and $$g(1) = f(2) ~- f(1) = f(0) ~- f(1) = -g(0).$$

If $$g(0) = 0$$, then $$f(0+1) ~- f(0) = 0$$, so $$f(1) = f(0)$$ and the solution is to take $$x_1 = 0$$ and $$x_2 = 1$$. If $$g(0) \neq 0$$, then $$g(1)$$ and $$g(0)$$ are nonzero numbers of equal magnitude but opposite sign. By the Intermediate Value Theorem, there exists $$c \in (0, 1)$$ such that $$g(c) = 0$$. Now the solution is to define $$x_1 = c$$ and $$x_2 = c+1$$. This makes $$g(c) = f(c+1) ~- f(c) = f(x_2) ~- f(x_1) = 0$$, so $$f(x_2) = f(x_1)$$. $$\blacksquare$$

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