without weak convergence, can strong convergence be true?

Question

I need help on part (b) for the linked question (please click on this link to see the question). I understand that the the omitted condition implies a weak convergence. Weak convergent alone does not imply strong convergent, however, together with the norm part in (a) can suggest strong convergent by showing the norm of $||f_n - g|| \rightarrow 0$. I believe the answer for (b) is "NOT," but I couldn't figure out how to either disapprove it or find a counterexample.

Thank you!

Answer 1

If $f$ and $g$ are orthogonal vectors of norm $1$ and $f_n=f$ for all $n$ then $\|f_n\| \to \|g\|$ but $f_n$ does not tend to $g$ in the norm.