Question: Consider $T_n : L^2[-\pi,\pi] \to L^2[-\pi,\pi]: u(x) \mapsto \displaystyle \int_{-\pi}^{\pi} \left( e^{\frac{x}{n} - \frac{y}{n}} \right) u(y) \; dy$, for $n \in \mathbb{Z}^{\ge 1}$.
(a) Is $T_n$ compact? Self-adjoint?
(b) Calculate $\|T_n\|_{op}$.
(c) Does $T_n$ converge? If so, to what and in what (operator) sense?
Solution:
(a) Compactness - Notice that $T_n (u) (x) = e^{\frac{x}{n}} \displaystyle \int_{-\pi}^\pi e^{-\frac{y}{n} } u(y) dy$, so ran$(T_n)$ is contained in the finite dimensional subspace spanned by $e^{x/n}$, meaning it's compact.
I calculated $T_n^*(u)(x) = \displaystyle \int_{-\pi}^{\pi} \left( e^{\frac{y}{n} - \frac{x}{n}} \right) u(y) \; dy$, so it's not self-adjoint.
(b) I'm stuck here. I can use Cauchy Schwarz to show that $\|T_n\|_{op} \le \|K\|_{L^2[-\pi,\pi]}$, where $K(x,y) = e^{x/n-y/n}$. But any attempts at the reverse inequality have not been successful . . .
(c) It converges to the identity . . . I think strongly. Because of LDCT I can push a limit inside the integral. Uniformly, I'm unsure.
This is an old test problem, so part of why I ask this question is because I will have to answer similar questions on the exam in a timely manner. I could maybe mess around with part (c) for a while and come up with some function $u(y)$ that demonstrates non-uniform convergence, but I'm curious if there's a better way to go about it. I can't think of any examples now.
Please let me know if my work is correct or you can offer help. Thank you
To calculate the norm we first make some heuristic argument. Looking at $T_n(u)(x)$, we see that this function is big, if the integral $\int_{-\pi}^{\pi}e^{-\frac{y}{n}}u(y)dy$ is big. When is this big? As $e^{-\frac{y}{n}}$ is big around $y = -\pi$, $u$ has to mass around $-\pi$. In the worst case, we have something like $\int_{-\pi}^{\pi}e^{-\frac{y}{n}}u(y)dy \leq \|u\|_{L^2} e^{\frac{\pi}{n}}$.
Now $||T_n(u)||_{L^2} = \int_{-\pi}^{-\pi}e^\frac{x}{n}|\int_{-\pi}^{\pi}e^{-\frac{y}{n}}u(y)dy|dx \leq \int_{-\pi}^{-\pi}e^\frac{x}{n}e^{\frac{\pi}{n}}\|u\|_{L^2}dx \leq \|u\|_{L^2} e^\frac{\pi}{n}2n \sinh{\frac{\pi}{n}}$.
The last $\sinh$ is just the $L^2$ norm of $e^\frac{x}{n}$. This formula is sharp by the heuristics, as one can take some sequence of $L^2$ functions converging to the $\delta$ distribution at $-\pi$.
For the third part, you have to consider that the identity is not compact, and norm convergence of compact operators are compact. Nevertheless we can pick some $u(x)$ which is orthogonal to $e^\frac{x}{n}$, then $\|T_n(u) - u\|_{L^2}\geq \|u\|_{L^2}$.
EDIT:
The limit is not the identity function. If $n\to \infty$, then $T_n(u)(x) \to \|u\|_{L^2}$, the constant function with value $\|u\|_{L^2}$. This is becuase $e^\frac{x}{n}$ converges uniformly to the constant $1$ function (at least on $[-\pi, \pi]$), and you can thus interchange limit and integral, without referencing LDCT.