Orthonormal basis on Hilbert space and strong operator topology In an infinite-dimensional Hilbert space $H$, let $(\phi_n)_{n=1}^{\infty}$ be an orthonormal basis. Let $A_n = (\phi_n, ·)\,\phi_1$. Prove that $A_n → 0$ in the strong operator topology, but that $A^∗_n$
has no limit in the strong operator topology.
In the problem attached above, I need to prove that $A_n \to 0$ in the strong operator topology.
So, I need to show that $||A_n(x)|| \to 0$ for all $x \in H$
$\implies ||(\phi_n,x)\,\phi_1|| \to 0$ $\implies ||(C_n)\,\phi_1|| \to 0.$ (As $x$ can be written as $\sum_{i}Ci\,\phi_i$.)
I have no idea how to proceed.
Also, please give hints for second part where I need to prove that $A_n^*$ has no limit in strong operator topology.
 A: (The syntax of what you've written is a bit garbled, I think... Best to write in sentences, come to a full stop, and then tell what is implied. Otherwise it's needlessly ambiguous...)
Yes, to show that $A_n\to 0$ in the strong operator topology, we must show that $|A_nx|\to 0$ for every fixed $x$. Now $A_nx=\langle \phi_n,x\rangle\cdot \phi_1$, and the norm of the latter is $|\langle \phi_n,x\rangle|$, since $|\phi_1|=1$. By Bessel's inequality, $\sum_n|\langle \phi_n,x\rangle|^2\le |x|^2$, so (since the infinite sum converges), $|\langle \phi_n,x\rangle|\to 0$, as desired.  (Here I suppose either that the Hilbert space is real, or that you're taking the convention that the pairing is complex linear in its second argument...)
Then, what is the adjoint of $A_n$?
EDIT: just so other people can see the rest of this... Changing some things by complex conjugation, which is inessential but convenient (especially for me to avoid telling a pointless lie!): making the inner product complex linear in its first argument, etc., let's characterize the adjoint $A_n^*$ by
$$
\langle x,A_n^*y\rangle\;=\; \langle A_n x,y\rangle
\;=\; \langle \langle x,\phi_n\rangle\cdot \phi_1,y\rangle
\;=\; \langle x,\phi_n\rangle\cdot \langle \phi_1,y\rangle
\;=\; \langle x,\langle y,\phi_1\rangle\cdot \phi_n\rangle
$$
By uniqueness of the adjoint, $A_n^*(y)=\langle y,\phi_1\rangle\cdot \phi_n$. Each of these image vectors has distance $\sqrt{2}\cdot |\langle y,\phi_1\rangle|$ from the others, which does not go to $0$ for $\langle y,\phi_1\rangle\not=0$. Thus, the sequence $A_n^*$ cannot be Cauchy in the strong operator topology, so cannot converge.
