hmmm16 said:
If I have the following operator for $H=L^2(0,1)$: $$Tf(s)=\int_0^1 (5s^2t^2+2)(f(t))dt$$ and I wish to calculate $||T||$, how do I go about doing this:
I know that in $L^2(0,1)$ we have that relation: $$||T||\leq \left ( \int_0^1\int_0^1 |(5s^2t^2+2)|^2dtds\right ) ^{\frac{1}{2}}=\sqrt{\frac{50}{6}}.$$
Check that you have done this correctly! I make the answer $\dfrac{\sqrt{65}}3$.
hmmm16 said:
I also have that $||T||=\sup_{||f||_2\leq1} ||Tf(s)||$ so if I take $f=1$ then this gives equality above and I am done, is that correct?
I do not understand that at all, and I don't believe it.
To find the operator norm of $T$, notice first that because the kernel $k(s,t) = 5s^2t^2+2$ is real-valued and symmetric in $s$ and $t$, it follows that $T$ is selfadjoint. Next, notice that, for any $f\in L^2(0,1)$, $$Tf(s) = \int_0^1 (5s^2t^2+2)(f(t))\,dt = s^2\int_0^15t^2f(t)\,dt + \int_0^12f(t)\,dt,$$ so that $T(f)$ lies in the two-dimensional subspace of $L^2(0,1)$ spanned by the functions $f(s) = s^2$ and $f(s) = 1$. Since $T$ is selfadjoint, its norm will be the norm of its restriction to that subspace.
So the strategy for finding $\|T\|$ is (1) use Gram–Schmidt to find an orthonormal basis for that subspace (I think that you can take the functions $f_1(s) = 1$ and $f_2(s) = \frac32\sqrt5\bigl(s^2-\frac13\bigr)$, but check that I got that right); (2) find what $T$ does to the two vectors in that basis, expressing $T(f_1)$ and $T(f_2)$ as linear combinations of $f_1$ and $f_2$.
That way, you reduce the problem to finding the norm of the $2\times2$-matrix of those coefficients. What's more, the matrix will be hermitian (because $T$ is selfadjoint), and the norm of the matrix will be equal to the size of its larger eigenvalue (in absolute value).
Operators given by integral kernels are always compact, and in this case the operator $T$ has finite rank 2. That is what makes it possible to find its norm explicitly.