# Show that $f(a)=f(a-1).$

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.

• You want $f(a)=f(a-1)$. So the most immediately obvious IVT function to try would be $f(x)-f(x-1)$. – Arthur Dec 18 '18 at 21:22
• 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... ) – Mason Dec 18 '18 at 21:24
• .Got it. Thank you! – gune Dec 18 '18 at 21:33

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.

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 $$g(a) = 0$$

• "if g(x)=0 you are done." should be g(1)=0 – Toby Bartels Dec 18 '18 at 21:29
• @TobyBartels thanks. – Doug M Dec 18 '18 at 21:29

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 $$x \in [0,2]$$.