Prove that there exists a value in [0,1] such that f(x_0)=g(x_0) for f:[0,1] to Reals and g:[0,1] to Reals.

Question

Suppose $f:[0,1]\to\Re$ and $g:[0,1]\to\Re$ are continuous, and their images both coincide with the closed unit interval: $f([0,1])=g([0,1])=[0,1]$. Prove that there exists ${x_0}\in[0,1]$ such that $f(x_0)=g(x_0)$.

So, I know I am going to be using the intermediate value theorem, but I am confused on how to get some actual values since there is technically no $f(x)$ or $g(x)$ defined. I did what I usually do in an IVT problem and I let $y(x)=f(x)-g(x)=0$. From here, I don't know which values to plug in to check if they are in between $[0,1]$ since I only have information about their image, continuity, and domain.

My professor suggested using a smaller interval between the minimizer and the maximizer of the function $f(x)$. But, how would I go about finding a minimizer and maximizer of a function that is not explicitly defined?

Answer 1

Suppose $f(x_1)=0$. If $g(x_1)=0$ then we are done. So we can assume $g(x_1)>0$, and thus your $y(x)=f(x)-g(x)$ is negative at $x_1$.

Similarly for $f(x_2)=1$. We can assume $g(x_2)<1$, and thus $y(x_2)>0$.

Since $y(x)$ is continuous, we know that $y(x_0)=0$ for some $x_0$ between $x_1$ and $x_2$.

Answer 2

If there is no such point then, by continuity of $f-g$, either $f>g$ or $f<g$ throughout $[0,1]$. In the first case, let $x$be a point where $g$ attains its maximum value $1$ Then $f(x)>1$, a contradiction. Similar argument works in the second case.