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If $H$ is Hilbert space, prove that $$ \|A\| \leq \liminf_{n \to \infty} \|A_n\|,$$ when $A$ is weak limit of $A_n$, for some $A_n \in B(H)$.

I know that $A_n$ converges weakly to $A$ iff $\lim\langle A_nf,g\rangle = \langle Af,g\rangle, \forall f,g \in H$. Also, $\|A\| = \underset{\|f\|=1}{\sup}|\langle Af,f\rangle|$. Further,

$$|\langle Af,f\rangle |=|\lim \langle A_nf,f\rangle| = \lim |\langle A_nf,f\rangle|.$$

Let $\varepsilon > 0$, arbitary. $\langle A_nf,f\rangle$ converges, so I can take subsequence $\langle A_{nk}f,f\rangle$, such that $$|\langle A_{nk}f,f\rangle| \leq \varepsilon + \liminf_{n \to \infty} \|A_n\|.$$ Since $\lim |\langle A_nf,f\rangle| = \lim |\langle A_{nk}f,f\rangle|$, I get the result.

I am not sure about part: $|\langle Af,f\rangle |=|\lim \langle A_nf,f\rangle| = \lim |\langle A_nf,f\rangle|$, because it's weak limit. Can someone tell me if this is wrong and help me if it is?

Thanks!

colt_browning
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Maria
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  • Thank you! Is my work correct? – Maria May 29 '23 at 15:12
  • the part you say you're unsure about is absolutely fine since its a (strong) limit of a sequence of real numbers in all cases (fixing $f$ with $|f|=1$). You might want to more carefully justify the inequality $|\langle A_{n_k}f,f\rangle|\le\epsilon+\liminf_n|A_n|$ – FShrike May 29 '23 at 18:25
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    Are you dealing with self-adjoint operators? Otherwise the formula $|A|=\sup_{|f|=1}|\langle Af,f \rangle |$ fails, in general: https://math.stackexchange.com/questions/298353/norm-of-bounded-operator-on-a-complex-hilbert-space – Gerd May 30 '23 at 06:57
  • No, not self-adjoint. Can I work with $<A_nf, g>$ instead? – Maria May 31 '23 at 16:38
  • @Maria Yes, $|A|=\sup_{|f|=1,|g|=1} |\langle Af,g \rangle |$ holds for each bounded operator $A$ – Gerd May 31 '23 at 17:17

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