I am given the following subset of $C([0,1])$ of continuous real-valued functions on $[0,1]$, satisfying $$ \lvert f(x) - f(y) \rvert < \lvert x - y \rvert, \quad \int_0^1 f(x)^2 dx = 1 $$ (I am assuming they meant to include the condition for $x \neq y$ in the above).
As a step in proving compactness using Arzela- Ascoli ( I have already shown uniform boundedness and equicontinuity), I am trying to prove the space is closed (with the sup norm), but having trouble with the strict inequality. I have shown that for a sequence $f_n \rightarrow f$, $$ \lvert f(x) - f(y) \rvert = \lvert \lim_{n\rightarrow \infty} (f_n(x) - f_n(y)) \rvert = \lim_{n\rightarrow \infty}\lvert f_n(x) - f_n(y) \rvert\leq \lvert x - y \rvert $$ but don't know how to prove the limit is strictly less than $\lvert x - y \rvert$. I was wondering if there is some contradiction of the equality using the second condition, or some other way to prove this, or if this is possibly a typo? Because to me it seems that if there were only the first condition, this is simply untrue since you could just define a sequence of functions that converged to $f(x) = x$, but maybe with the second condition it is possible? I appreciate any assistance with this question.
That doesn't seem to be true.
Let $$f_{a, b}(x) = \begin{cases} a + 3x(1 - a) & x \leq \frac{1}{3}\\ 1 & \frac{1}{3} < x \leq \frac{2}{3}\\ 1 + 3(x - \frac{2}{3})(b - 1) & \frac{2}{3} < x \end{cases}$$
Plot of it is chain $(0, a) \to (1/3, 1) \to (2/3, 1) \to (1, b)$. The first condition is satisfied if $a > \frac{2}{3}$, $b < \frac{4}{3}$.
We can compute integral explicitly, but it's clear that there is some monotonic function $g$ s.t. then $\int_0^1 f^2_{a, g(a)}(x)^2\, dx = 1$. Now, if $g(2/3) \leq 4/3$, consider sequence $f_{2/3 + 1/n, g(2/3 + 1/n)}$ - it is in your subset, but converges to $f_{2/3, g(2/3)}$, which is not.
If $g(2/3) > 4/3$ then $g^{-1}(4/3) > 2/3$, so similarly consider sequence $f_{g^{-1}(4/3 - 1/n), 4/3 - 1/n}$.