Norm of Fredholm operator in $L^1$ Let $T:L^1([0,1])\rightarrow L^1([0,1])$ be the Fredholm integral operator given by
$$
Tf(x)=\int_0^1 k(x,y)f(y)\, dy
$$
where $k \in C([0,1]^2)$ is called the kernel of $T$. My problem is to find $\|T\|$ in terms of it's kernel. 
It's easy to show that it is bounded, but I'm failing to prove that the bound it is the norm (using sign functions). I know a similar result for $T$ acting on $C([a,b])$, a bound for $T$ acting in $L^2$, (and, more generally, from $L^p$ to $L^q$), but here I need help in this case.
 A: I don't see why the following doesn't work: Let $T$ be the Fredholm operator on any Banach space, call it $X$. By definition, $||T||_X = \sup_{|f|\leq 1}|Tf|_X$, so
$$||T||_X = \sup_{|f|_X\leq 1}\left|\int_0^1k(x,y)f(x)dx\right|_X\leq \sup_{|f|_X\leq 1}\int_0^1|k(x,y)|_X|f(x)|_Xdx = \int_0^1|k(x,y)|_Xdx.$$
A: Here is my solution. The idea is to not to find a function satisfying $||T|| = \frac{||Tf||_{L^1[0,1]}}{||f||_{L^1[0,1]}}$, but to find a sequence of functions such that $||T||=\lim\limits_{n \to \infty} \frac{||Tf_n||_{L^1[0,1]}}{||f_n||_{L^1[0,1]}}$.
Using Fubini's theorem and Hölder's inequality it's easy to see:
$$||Tf||_{L^1[0,1]} = \int_0^1|Tf(y)|dy = \int_0^1\left|\int_0^1k(x,y)f(x)dx\right|dy \le \int_0^1\int_0^1|k(x,y)||f(x)|dxdy = \int_0^1 dx|f(x)|\int_0^1 dy|k(x,y)| \le ||f||_{L^1[0,1]} \left|\left|\int_0^1 |k(x,y)|dy\right|\right|_{C[0,1]} $$
So the guess is that $||T|| = \left|\left|\int_0^1 |k(x,y)|dy\right|\right|_{C[0,1]}$.
$\DeclareMathOperator*{\argmax}{arg\,max}$
Consider $f_n(x)=I_n(x)$ $-$ indicator of $U_{\frac{1}{2n}}(x_0)\cap[0,1]$, where $x_0 = \argmax_\limits{x \in [0,1]}\int_0^1 |k(x,y)|dy$.
Using integral mean value theorem:
$$||T||=\lim\limits_{n \to \infty} \frac{||Tf_n||_{L^1[0,1]}}{||f_n||_{L^1[0,1]}} = \lim\limits_{n \to \infty} \frac{\int_0^1\left|\int_{x_0-\frac{1}{2n}}^{x_0+\frac{1}{2n}}k(x,y)dx\right|dy}{\frac{1}{n}} = \lim\limits_{n \to \infty}\frac{\int_0^1|\frac{1}{n}k(\xi_n,y)|dy}{\frac{1}{n}} = \int_0^1|k(x_0,y)|dy = \left|\left|\int_0^1 |k(x,y)|dy\right|\right|_{C[0,1]} $$
