What is the Density Theorem in this context?

Question

I have this exercise:

Define $K:C([0,1])\rightarrow C([0,1])$ by $$Kf(x) = \int_0^1 k(x,y)f(y)dy,$$ where $k:[0,1]\times [0,1]\rightarrow \mathbb{R}$ is continuous. Prove that $K$ is bounded and $$||K|| = \max_{0\leq x\leq 1}\bigg\{\int_0^1|k(x,y)|dy\bigg\}.$$

This is what I have so far: Let $A = \max_{0\leq x\leq 1}\bigg\{\int_0^1|k(x,y)|dy\bigg\} = \max_{0\leq x\leq 1}\bigg\{\int_0^1|k(x_0,y)|dy\bigg\}$.

I want to show that $||K||\geq A$. I have this proof,

Let $$ f_{\infty}(x) = \begin{cases} 1 & k(x_0,y)>0 \\ -1 & k(x_0,y)<0 \\ 0 & \text{otherwise} \end{cases}$$ Then, $\int_0^1 k(x_0,y)f_{\infty}(y)dy = \int_0^1 |k(x_0,y)|dy = A$.

By the $\bf{DENSITY THEOREM}$, there exists $f_n\in C([0,1])$ such that $||f_n||=1$ and $f_n(x)\rightarrow f_{\infty}(x)$ almost everywhere.

$\implies$ $||K||\geq\dfrac{||Kf_n(x)||}{||f_n(x)||}\geq |Kf_n(x_0)| = |\int_0^1 |k(x_0,y)f_n(y)dy| \rightarrow |\int_0^1 |k(x_0,y)f_{\infty}(y)dy| = A$

What is the Density Theorem?

Answer 1

The function $f_\infty$ is integrable and the continuous functions on the unit interval are dense in $L^1$. So we approximate $f_\infty$ in $L^1$ by continuous functions. With a fast enough approximation (say $\lVert f_\infty-f_n\rVert_{L^1}\leqslant 2^{-n}$), we get almost everywhere convergence.