# Fredholm operator norm

I have seen here, that the operator norm of a Fredholm operator $T_k(f)(s):=\int_0^1 k(s,t) f(t) dt$, where $k \in L^2([0,1]^2)$ and $f \in L^2([0,1])$ is not equal to the $L^2$ norm of the Kernel.

But can I show that for $T_k:C(a,b)\rightarrow C(a,b)$ and $k\in C([a,b]^2)$ $$\Vert T_k\Vert=\sup_{x\in [a,b]}\int_a^b|k(x,y)|dy\;?$$

#### Solutions Collecting From Web of "Fredholm operator norm"

One direction is easy: just take supremum over $s$ on both sides of
$$|T_kf(s)|\le \int_a^b |k(s,t)| |f(t)| \,dt \le \|f\|_{C[a,b]} \int_a^b |k(s,t)| \,dt$$

The converse direction is also easy when $k\ge 0$, because we can take $f\equiv 1$ to immediately obtain
$$T_kf(s) = \int_a^b k(s,t) \,dt$$
for all $s$.

But in general case we have to work harder. It’s a little easier for real-valued functions, but I’ll go with the complex case (which also works for real case).

Fix $s_0$ that achieves the supremum of $\int_a^b |k(s,t)| \,dt$. The function $g(t) = \operatorname{sign} k(s,t)$ is in $L^1[a,b]$ and therefore can be approximated in $L^1$ norm by a continuous function $h$. Say, $\|h-g\|_{L^1}<\epsilon$. Moreover, we can make sure $|h|\le 1$ pointwise by truncation: when $|h(t)|>1$, define $\tilde h(t) = h(t)/|h(t)|$; otherwise $\tilde h(t)=h(t)$. It’s an exercise to check that $$|\tilde h(t)-g(t)|\le |h(t)-g(t)|$$ pointwise. Consequently, $\|\tilde h-g\|_{L^1}<\epsilon$. Finally, estimate $T_k\tilde h(s_0)$:
$$|T_k\tilde h(s_0)|\ge |T_k g(s_0)| – \epsilon \sup|k| = \int_a^b |k(s_0,t)| \,dt – \epsilon \sup|k|$$
Since $\epsilon$ was arbitrarily small, and $\|\tilde h\|_{C[a,b]}\le 1$, the claim follows.