Norm of Fredholm integral operator equals norm of its kernel? Let $T_k(f)(s):=\int_0^1 k(s,t) f(t) dt $, where $k \in L^2([0,1]^2)$ and $f \in L^2([0,1])$.
Then it was fairly easy to see that $||T_k|| \le ||k||_{L^2}$, but now I was wondering how to show that we have $||T_k|| = ||k||_{L^2}$, or is there a lower bound?
 A: I don't think it is true that $\|T\|_{L^2[0,1]} = \|k(s,t)\|_{L^2([0,1]^2)}$. We will show that for $k(s,t) = \chi_{[0,s]}(t)$, $\|T\|_2 = 2/\pi \approx 0.63$ which is strictly less than 
$$ \|k(s,t)\|_2 = \sqrt{ \int_0^1 \int_0^1 \chi_{[0,s]}(t) dt ds } 
= \sqrt{\int_0^1 s ds}
= \frac{1}{\sqrt{2}} \approx 0.7. $$ Note that when $k(s,t)$ is as above, $$(Tf)(s) = \int_0^1 \chi_{[0,s]}(t) f(t) dt = \int_0^s f(t) dt.$$ We need the following facts:


Lemma 1: Let $T$ a compact self-adjoint operator on a Hilbert space $\mathcal{H}$. Then $$\|T \| = \max_{\lambda \text{ an eigenvalue of $T$}} |\lambda|.$$
    Proof: We recall a result from spectral theory which states that $\|T\| = \max_{\lambda \in \sigma(T)} |\lambda|$, where $\sigma(T)$ is the spectrum of $T$. Then since every eigenvalue is in the spectrum, we have
    $$\begin{eqnarray*} \max_{\lambda \text{ an eigenvalue of $T$}} |\lambda| &\leq& \max_{\lambda \in \sigma(T)} |\lambda| \\
&=& \|T\|.\end{eqnarray*}$$
    On the other hand, by Lemma 6.5, chapter 4 of Stein and Shakarchi volume 3, we have that either $-\|T\|$ or $\|T\|$ is an eigenvalue of $T$, showing the reverse inequality. This proves the lemma.





Lemma 2: Let $T$ be a bounded operator on a Hilbert space $\mathcal{H}$. Then 
    $$\|TT^\ast\| = \|T^\ast T\| = \|T\|^2 = \|T^\ast\|^2.$$
    Proof: Clear from the fact that for a self-adjoint operator $U$, $\|U\| = \sup_{\|x\| = 1} | \langle Ux,x\rangle|$.



We now come to the main proposition which is crucial to our result. Let $T$ be as defined above. Then note $T$ is a compact operator because it has $L^2([0,1]^2)$ kernel. We prove:


The largest eigenvalue of $T^\ast T$ is  $4/\pi^2$.


Proof: Suppose $T^\ast Tf = \lambda f$. We may suppose $\lambda \neq 0$ for we are trying to find the largest $\lambda$, and a corollary of the Spectral Theorem is that a compact self-adjoint operator always has at least one non-zero eigenvalue. Now by direct computation we find the equation $T^\ast T = \lambda f$ translates to
$$\int_1^t \int_0^s f(t) dt ds = \lambda f(t).$$
Using the fundamental theorem of calculus twice, we get $\lambda f'' + f = 0$ subject to the boundary conditions $f(1) = 0$ and $ f'(0) = 0$. Now the differential equation $f'' + f/\lambda = 0$ has solution
$$f(t) = Ae^{\mu t} + Be^{-\mu t}$$
where $\mu = \sqrt{-1/\lambda}$. The latter boundary condition gives $A = B$, while the former shows $e^\mu + e^{-\mu} = 0$ and so $e^{2\mu} = -1$. Taking principal logs, we get $2\mu = i \pi  + 2k\pi i = (2k + 1)i\pi $ and so 
$$\lambda = \frac{4}{(2k+1)^2\pi^2}.$$
The largest such $\lambda$ is when $k = 0$, so Lemma 1 shows that $\|T^\ast T\| = 4/\pi^2$. It follows from Lemma 2 that $\|T\| = 2/\pi$. This shows it is not always the case that the norm of the Fredholm operator is the norm of its kernel.
