Find a norm of integral operator on the space of continuous functions I know this question has been asked  here and thank you for the answer. But there is some detail that I don't get it. And I don't have enough reputation to comment.
Let's consider $C[0,1]$ endowed with uniform norm ${\parallel f\parallel}_{\infty} =\displaystyle\sup_{t\in [0,1]}\|f(t)|$.
Let $T:C[0,1]\rightarrow C[0,1]$ defined by:
$$ Tf(t) = \int_{0}^{1} K(t,s) f(s) ds $$ where $K\in C([0,1]\times[0,1])$.
Show that $||T|| = \displaystyle\sup_{t\in [0,1]} {\int_{0}^{1} |K(t,s)| ds} $
@roo gave the answer quite clearly, but the part when choosing a sequence $(g_n)$ in $C[0,1]$ which converges to $g :=\text{sign}[K(t_{0},\cdot)]$. How you guarantee that such a sequence exists?
Thanks.
 A: It is guaranteed by the following two facts:
$1$. The continuous functions on $[0,1]$ are dense in $L_1[0,1]$.
$2$. If $g_n\to g$ in $L_1$, then there is a subsequence of $g_n$ converging to $g$ almost everywhere.
(The measure is the usual Lebesgue measure).
These are standard measure-theoretic facts. You will find them in almost every textbook. For example, in Rudin's 'Real and Complex Anaylsis'.
A: In the answer of @roo:, he considers $t_0\in [0,1]$ such that
$$\int_0^1 |K(t_0, s)| = \max_{t \in [0,1]} \int_0^1 |K(t,s)| ds$$
(the function of $t$ on the RHS is continuous, not hard to check).
Now, consider the continuous function $s \mapsto K(t_0, s)$. The subsets  $U = \{ s\in [0,1]\ |\  K(t_0, s) < 0\}$, and $V = \{s \ |\  K(t_0, s) > 0\}$ are open in $[0,1]$. Take any $\epsilon > 0$.  Consider now $C$, $D$ compacts subsets of $U$, $V$ such that
$$\int_{C\cup D} |K(t_0, s)|\, ds> \int_0^1 |K(t_0, s)|\,ds - \epsilon$$
($C$, $D$ are finite unions of closed intervals).
Now take $f_1\colon [0,1] -> [0,1]$, continuous, $f= 1$ on $C$, $0$ outside $U$, and $f_2 = 1$ on $D$, $0$ outside $V$ ( this is fairly standard, say $f_1(t) = \frac{d_C(t)}{d_C(t) + d_{U^c}(t)}$), and $f = f_1 - f_2$. You got an $f$ of norm $1$ $\ldots$.
