I have a set of functions given by; $$F = \{f:[0,1]\rightarrow\mathbb{R}|\int_0^1 f(x)dx = 0, |f(x)-f(y)|\leq|x-y|, x,y\in[0,1]\}.$$ I have a solution for the question so my questions are about the solution.
Let $f\in F \implies \int_0^1 f(x)dx = 0$.
$\implies$ there exists $x_0\in [0,1]$ such that $f(x_0) = 0$.
$\implies |f(x)| = |f(x)-f(x_0)|\leq |x-x_0|\leq 1$.
Therefore $F$ is bounded.
My questions are:
How can we deduce that there exists an $x_0\in [0,1]$ such that $f(x_0)=0$?
How do we see that $|x-x_0|\leq 1$? I know the Lipschitz constant is less than or equal to one but how does that imply $|x-x_0|\leq 1$?
Thank you for any help and comments.
Since $|f(x) - f(y)| \le |x - y|$, $f$ is continuous (and Lipschitz continuous with constant at most $1$) as you've noted. By the mean value theorem, such an $x_0$ must exist: The MVT for integrals states that for some $x_0$,
$$f(x_0) = \frac{1}{1 - 0} \int_0^1 f(t) dt = \frac{0}{1}$$
Then since $x_0 \in [0, 1]$ and $x \in [0, 1]$, it is immediate that $|x_0 - x| \le 1$; they are elements of the same interval of length $1$.