prove the following statement:
Assume $f(x)$ is continuous over $\left[-1,1\right]$ and $∀x∈\left[-1,1\right]$ :$\left|f\left(x\right)\right|\le1$, also assume $g(x)$ is continuous over $\left[-1,1\right]$ and $g(-1)=- 1$ , $g(1)=1$.
Prove $$∃x_{0}∈\left[-1,1\right]:f\left(x_{0}\right)=g\left(x_{0}\right)$$
My try:
Define : $$h\left(x\right):=f\left(x\right)-g\left(x\right)$$
Then $$h\left(1\right)=f\left(1\right)-g\left(1\right)=f\left(1\right)-1\le0$$
Also $$h\left(-1\right)=f\left(-1\right)-g\left(-1\right)=f\left(-1\right)+1\ge0$$
Since $f(x)$ and $g(x)$ are continuous over $\left[-1,1\right]$, implies $h(x)$ is continuous over that interval, and since $h\left(-1\right)h\left(1\right)\le0$, using Intermediate value theorem implies
$$∃x_{0}∈\left[-1,1\right]:h\left(x_{0}\right)=0$$
Or equivalently $$∃x_{0}∈\left[-1,1\right]:f\left(x_{0}\right)=g\left(x_{0}\right)$$
To be absolutely rigorous, I'd just highlight a few things.
If $h(1) = 0$ or $h(-1) = 0$, we are done. Otherwise, we have $h(-1) > 0$ and $h(1) < 0$. Then, by the Intermediate Value Theorem, there is some $x \in (-1, 1)$ with $h(x) = 0$.
Whatever the case, there is some $x\in[-1,1]$ with $h(x) = 0$, that is, with $f(x) = g(x)$.
What's important to highlight is that the intermediate value theorem grabs some $x$ strictly in between the endpoints, and that the value also lies strictly in between the value at the endpoints.