Operator Convergence in Hilbert space

Question

I'm wondering if you can check my work and provide help on part of this problem. Here's the question: Let $H$ be Hilbert and $(\phi_n)$ an orthonormal basis. Let $(a_j)$ be a sequence of nonnegative reals such that $a_j \to \infty$. Define $T_n(t) \in B(H)$ by $$ T_n(t)[u] = \sum_{j=1}^n e^{- a_j t} \langle \phi_j, u \rangle \phi_j $$ (a) Show that $(T_n(t))_{n=1}^\infty$ converges to some $T(t) \in B(H)$ for any fixed $t \in [0,\infty)$. What is the type of convergence?

(b) Does the limit $T(t)$ converge as $t \to 0$? what is the type of convergence?

Here's my work: It is clear that $\sum_{j=1}^\infty e^{- a_j t} \langle \phi_j, u \rangle \phi_j$ is a bounded linear operator: $$ \left\| \sum_{j=1}^\infty e^{- a_j t} \langle \phi_j, u \rangle \phi_j \right\|^2 = \sum_{j=1}^\infty | e^{-2 a_j t} | \langle \phi_j, u \rangle|^2 \le \sum_{j=1}^\infty | \langle \phi_j, u \rangle|^2 = \|u\|^2 < \infty $$ and linearity is trivial. Also $T_n(t) \to T(t)$ uniformly. Consider $\|u\|=1$, then $$ \|T(t) - T_n(t) \|_{op} \le \left\| \sum_{j=n+1}^\infty e^{- a_j t} \langle \phi_j, u \rangle \phi_j \right\|^2 = \sum_{j=n+1}^\infty |e^{- a_j t}|^2 | \langle \phi_j, u \rangle|^2 \le \sum_{j=n+1}^\infty | \langle \phi_j,u \rangle|^2 \stackrel{n \to \infty}{\longrightarrow} 0 $$

For part (b), from LDCT we have $\lim \limits_{ t \to 0} \sum_{j=1}^\infty e^{- a_j t} \langle \phi_j ,u \rangle \phi_j = \sum_{j=1}^\infty \langle \phi_j, u \rangle \phi_j = u$ so $T(t) \to I$ as $t \to 0$.

But I'm uncertain about the type of convergence here since operator, strong, and weak convergence were defined for sequences of operators... not something like this.

Does my work look OK for (a) and part of (b)? Also, if you see a better/slicker way to get these results let me know. Studying for an exam.

Answer 1

The first inequality in your operator norm convergence does not hold (an operator does not necessarily achieve its norm, and even if it did your $u$ would depend on $n$ and $t$). But you have \begin{align} \|[T(t)-T_n(t)]u\|^2&=\|\sum_{j=n+1}^\infty e^{-a_jt}\,\langle \phi_j,u\rangle\,\phi_j\|^2=\sum_{j=n+1}^\infty e^{-a_jt}\,|\langle\phi_j,u\rangle|^2\\ \ \\ &\leq \sum_{j=n+1}^\infty |\langle\phi_j,u\rangle|^2\xrightarrow[n\to\infty]{}0. \end{align} This shows strong operator convergence. When $t=0$, you cannot do better than that, because $\|T(0)-T_n(0)\|=1$ (it is a projection) for all $n$.

When $t>0$, the exponentials play a role. We have \begin{align} \|[T(t)-T_n(t)]u\|^2&=\|\sum_{j=n+1}^\infty e^{-a_jt}\,\langle \phi_j,u\rangle\,\phi_j\|^2=\sum_{j=n+1}^\infty e^{-a_jt}\,|\langle\phi_j,u\rangle|^2\\ \ \\ &\leq \|u\|^2\,\sum_{j=n+1}^\infty e^{-a_j t}. \end{align} As $u$ is arbitrary, the above shows that, for $t>0$, $$ \|T(t)-T_n(t)\|_{\rm op}^2\leq \sum_{j=n+1}^\infty e^{-a_j t}. $$ Since $a_n\to\infty$ the series is convergent, and so $T_n(t)\to T(t)$ in norm.

Regarding $T(t)$, your argument shows that $T(t)\xrightarrow[t\to 0]{}I$ in the strong operator topology. Note that \begin{align} \|[T(t)-I]\phi_k\|^2&=\sum_{j=1}^\infty (1-e^{-a_j t})\,|\langle \phi_j,\phi_k\rangle|^2=1-e^{-a_kt}, \end{align} which implies that $\|T(t)-I\|_{\rm op}=1$ for all $t>0$.