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Suppose that $f :[0,2]\rightarrow \mathbb{R}$ is a continuous function with $f(0)=f(2)$. Show that there is a real number $a\in[1,2]$ with $f(a)=f(a-1)$.

For the answer I tried to apply the mean value theorem to the function $g(x)=\int_0^2f(t)dt$ in the interval $[0,2]$ but it didn't lead any where.

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    $\begingroup$ You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$. $\endgroup$ – Arthur Dec 18 '18 at 21:22
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    $\begingroup$ Define $g(x)=f(x)-f(x-1)$, Then consider $g$ at $x=1$ and $x=2$ $g(1)=f(1)$ $g(2)=-f(1)$ so they have opposite sign... (well... if they don't have opposite sign then it's even easier... ) $\endgroup$ – Mason Dec 18 '18 at 21:24
  • $\begingroup$ .Got it. Thank you! $\endgroup$ – gune Dec 18 '18 at 21:33
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Define a function $g\colon [1,2] \to \mathbb{R}$ by $g(x) = f(x) - f(x-1)$, so that you're looking for an $a$ such that $g(a) = 0$. Check that $g$ is continuous and apply the Intermediate Value Theorem.

You need to know something about $f(1)$ to do this properly, but you can consider separately the three cases where $f(1)$ is greater than, less than, or (the trivial case) equal to $f(0)$ and $f(2)$.

ETA: While I was writing this, two comments appeared suggesting the same approach.

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

$g(2) = f(2) - f(1) = f(0) - f(1) = -g(1)$

if $g(1) = 0$ you are done.

Otherwise $g(x)$ is a continuous function that is negative at one endpoint and positive at the other.

By the IVT there must be an $a\in [1,2]$ such that $g(a) = 0 $

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  • $\begingroup$ "if g(x)=0 you are done." should be g(1)=0 $\endgroup$ – Toby Bartels Dec 18 '18 at 21:29
  • $\begingroup$ @TobyBartels thanks. $\endgroup$ – Doug M Dec 18 '18 at 21:29
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Elementary Approach using Bolzano's Theorem and not IVT :

Let $g(x) = f(x) - f(x-1)$. This function is continuous in $[0,2]$ as an operation of continuous functions.

Now, $g(1) = f(1)- f(0)$ and $g(2) = f(2)-f(1)$.

$$g(1)\cdot g(2) =f(1)f(2)-f(1)^2 - f(2)f(0)+ f(0)f(1)=2f(0)f(1) - f(1)^2-f(0)^2 $$ $$\implies$$ $$g(1)\cdot g(2) = -\big[f(1)-f(0)\big]^2 \leq 0 $$

Thus, there exists $a \in [0,2]$ such that $g(a) =0$ by Bolzano's Theorem.

Functional Analysis (sorry for this) :

Essentialy, we want to show that the equation

$$f(x) - f(x-1) =0$$

has a solution.

Let $T$ be a linear operator $T:C[0,2] \to C[0,2]$ such that $Tf(x) = -f(x-1)$. One can easily see that $T$ is a bounded linear operator $\in B\big(C[0,2]\big)$. Also, one can take $0$ to be a function defined in $C[0,2]$. Then, the equation is re-written as :

$$f + Tf = 0$$

Note that $C[0,2]$ is a Banach Space and since $T \in B\big(C[0,2]\big)$ then there exists a unique solution in $C[0,2]$, meaning that there exists such function $f(x) \in C[0,2]$ which of course implies that the equation has a solution for $x \in [0,2]$.

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