Let $K := \{x \in C[0,1] : x(0) \in [-3,4], |x(t)-x(s)| \leq d |t^2-s^2|, \forall t,s \in C[0,1]\}$. Let $y \in C[0,1]$ and $f : K \to \mathbb{R}$ defined as $f(x)= \int_0^1 x(t)y(t)dt$. Show that $f$ is uniformly continuous.
Partial proof :
- $f(x)= \int_0^1 x(t)y(t)dt$ is a dot product, noted $<x,y>$.
- $||f||_2 = \sqrt{<x,y>}$ is a continuous function.
- $g \circ ||f||_2$ is a continuous function, where $g : \mathbb{R} \to \mathbb{R}$ difined as $g(x)= x^2$
- $C[0,1]$ is compact
Thus it is sufficient to show that $K$ is closed (well-known theorem).
Is it hard to show that $K$ is closed? Am I on the good way? Is there a simpler way to proof this problem?
Edit :
Theorem : Let ${x_n}$ a sequence in $C[0,1]$ such that
- for all $t \in [0,1]$, the sequence ${x_n(t)}$ is bound over $\mathbb{R}$
- the set ${x_n : n \in \mathbb{N}}$ is equicontinuous.
Then there exists a subsequence $\{x_{n_k}\}$ uniformly convergent.
The teacher of the course left a theorem seen in a previous course, is that this theorem can be helpful in this question?
Hint: This is an alternative to your approach that's more straightforward. For all $x \in C[0,1]$ (hence, for all $x \in K$) $$ |f(x)| \leq \|x\|_\infty\|y\|_\infty. $$ (Why? Show this if you don't have a theorem that proves this already.) Hence $$ \|x_n\|_\infty \to 0\text{ as }n \to \infty \implies |f(x_n)| \to 0\text{ as }n\to\infty, $$ as well.