If $f_n \to f$ in $\|\cdot\|$, then $\|f_n\| \to \|f\|$. When is the converse true?

Question

Suppose $f_n,f$ are elements in some normed space, and $f_n \to f$ in some norm $\|\cdot\|$, then by the reverse triangular inequality $\|f_n\| \to \|f\|$ is automatic. Are there a number of useful results which say when the converse is true; that is, if $\|f_n\| \to \|f\|$, then $f_n \to f$ in some norm $\|\cdot\|$?

It is clearly not usually true, for if $z_n$ and $x_n$ are sequences of complex numbers such that $z_n \to e^{i\pi/2}$ and $x_n \to 1$, then they converge in norm on $\mathbb{C}$, but $z_n \not \to x_n$ in $\mathbb{C}$.

But are such statements (1) true in some of the useful norms, or (2) does there exist any useful results with regards to this question? If the answer is indeed no; I accept answers with insight which can help to understand why it is the case.

Answer 1

As noted by @Severin Schraven, there is this result for Hilbert spaces:

Let $H$ be a Hilbert space, $(f_n)_n$ a sequence in $H$ and $f \in H$. Then $$f_n \xrightarrow{\|\cdot\|} f \quad \iff \quad \|f_n\| \xrightarrow{} \|f\| \,\,\text{ and } \,\, f_n \xrightarrow{w} f$$

Proof:

If $f_n \xrightarrow{\|\cdot\|} f$ then also $f_n \xrightarrow{w} f$ and clearly $\|f_n\| \xrightarrow{} \|f\|$ because the norm is continuous.

Conversely, if $\|f_n\| \to \|f\|$ and $f_n \xrightarrow{w} f$ then

$$\|f - f_n\|^2 = \|f_n\|^2 - 2\operatorname{Re}\langle f, f_n\rangle + \|f\|^2 \to \|f\|^2 - 2\operatorname{Re} \langle f, f\rangle + \|f\|^2 = 0$$ so $f_n \xrightarrow{\|\cdot\|} f$.