How do I prove this set is completely bounded?

Question

Let $K \in C([0,1]\times[0,1])$. For $f \in C([0,1])$, define:

$$Tf(x) = \int_{0}^{1} K(x,y)\ f(y) \ dy, \ \ \ \ x \in [0,1]$$.

With $||f||_{\infty} := sup_{x \in [0,1]} |f(x)|$ norm, prove that the set

$$\{Tf : f \in C ([0,1]), \ ||f||_{\infty} \le 1 \}$$

is completely bounded in $C([0,1])$.

I tried using Arzela-Ascoli but couldn't prove it.

Answer 1

You mean "totally bounded" as in "can be covered by finitely many subsets of any fixed 'size'". Which is a given if the set is pre-compact. Whence the use of Arzela-Ascoli makes sense.

As \begin{align} |(Tf)(y)-(Tf)(x)|&\le\int_0^1|K(y,s)-K(x,s)|·|f(s)|\,ds\\ &\le\int_0^1|K(y,s)-K(x,s)|\,ds \end{align} the uniform continuity of $K$ on the compact set $[0,1]^2$ translates into the equi-continuity of the image of the unit ball.

Answer 2

Let $M$ be the maximum of $|K|$ on $[0,1] \times [0,1]$. For $f$ with $||f||_{\infty} \le 1$ we get

$|(Tf)(x)| \le \int_0^1 M ||f||_{\infty} dx =M ||f||_{\infty} \le M$ for all $x \in [0,1]$

Thus: $ ||Tf||_{\infty} \le M$