Below I am trying to prove problem 5.10 from Hunter's book $\textit{Applied Analysis}$ from UC Davis. The problem asks to prove that the integral operator $K$ is compact given that $k$ is continuous. I am concerned that I am choosing the wrong Cauchy Sequence $(f_n)$. Does $(f_n)$ need to be Cauchy or do I need to choose a sequence such that $(Kf_n)$ is Cauchy? I believe that since $K$ represents a linear operator it maps Cauchy sequences to Cauchy sequences correct? In which case the assumption that $(f_n)$ is Cauchy should be fine. If someone could read through my proof and provide suggestions I would greatly appreciate it.
Suppose $k : [0,1] \times [0,1] \to \mathbb{R}$ is a continuous function.
Let $K: C([0,1]) \to C([0,1])$ defined by $$Kf(x) = \int_0^1 k(x,y)f(y)dy$$ is compact. Let $\epsilon >0$
Since $k: [0,1] \times [0,1] \to \mathbb{R}$ is continuous, it is integrable.
So $\int_0^1 |k(x,y)|dy = M < \infty$ for some $M \in \mathbb{R}$ and for all $(x,y) \in [0,1] \times [0,1]$.
Now consider the Cauchy sequence, $(f_n) \in C([0,1])$.
We know that $C([0,1])$ is a Banach Space, which means $\exists n_0 \in \mathbb{N}$ s.t. $\forall n> n_0$ $||f_n - f|| < \frac{\epsilon}{M}$
Now for $n > n_0$ consider
$||Kf_n(x)-Kf(x)||$ = $||\int_0^1 k(x,y)f_n(y)dy - \int_0^1 k(x,y)f(y)dy||$
= $||\int_0^1 k(x,y)(f_n(y)-f(y))dy||$
= $\int_0^1 ||k(x,y)|| ||f_n(y)-f(y)||dy$
< $\frac{\epsilon}{M} \int_0^1 ||k(x,y)||dy$
= $\frac{\epsilon}{M} M = \epsilon$
So $K$ is compact.
A function between metric spaces maps Cauchy sequences to Cauchy sequences if and only if it is uniformly continuous. Linear maps between normed spaces are uniformly continuous if and only if they are continuous.
A (necessarily continuous) linear operator $f:X\to Y$ is said to be compact if it maps the (open or closed it makes no difference) unit ball of $X$ to a relatively compact subset of $Y$. Equivalent restatements of this:
if and only if $f$ maps the unit ball to a totally bounded subset of $Y$
if and only if for all sequences (not necessarily Cauchy) $x_n\in X$ such that $\lVert x_n\rVert<1$ there is a subsequence $x_{n_k}$ such that $f\left(x_{n_k}\right)$ is Cauchy in $Y$.
So, for your specific problem, you can't quite make the assumption that the $x_n$-s are Cauchy, because you must find Cauchy subsequences in the image for all sequences. The big hint for this is using Ascoli-Arzelà theorem.