$L^2$ operator norm and convergence 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
 A: 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.
