### Question


Let $X$ be some reflexive Banach space. Suppose $x\_n$ is some sequence in $X$ that weak converges to some $y \neq 0$. Is it the case that
$$ \limsup \|x\_n - y\| < \limsup \|x\_n\| ?$$


### Positive with Hilbert spaces


In the Hilbert space case this is true, as
$$ \langle x\_n - y, x\_n -y \rangle = \| x\_n\|^2 + \|y\|^2 - \langle x\_n, y\rangle - \langle y, x\_n\rangle $$
and by weak convergence the latter two terms both converges to $\|y\|^2$ and we get
$$ \limsup \|x\_n - y\|^2 = \limsup \|x\_n\|^2 - \|y\|^2. $$


### A negative example with a non-reflexive Banach space and weak\* convergence


Let $X = L^\infty(\mathbb{R})$, and take $x\_n = \mathbf{1}\_{[-2,-1]} + \mathbf{1}\_{[n,n+1]}$. Then $x\_n$ weak\* converges to $\mathbf{1}\_{[-2,-1]}$ but $\|x\_n - \mathbf{1}\_{[-2,-1]}\|\_{\infty} = 1 = \|x\_n\|\_\infty$.



Subtracting the weak limit reduces the norm in the limit
The property you indicate is known as (strict) Opial’s Property (see <https://en.m.wikipedia.org/wiki/Opial_property>). It fails generally in reflexive spaces; in fact, it fails generally even for uniformly convex spaces where it is equivalent to Delta convergence (see <https://en.m.wikipedia.org/wiki/Delta-convergence>). Indeed, for $L^p[0,1]$, where $p\ne 2$, Opial’s Property fails. Using the duality pairing $\langle\cdot\,,\cdot\rangle\colon X\times X^\*\to \mathbb{R}~or~\mathbb{C}$ and (the normalized) duality map $$J(x):=\{f\in X^\*:\|f\|^2=f(x)=\|x\|^2\}\,,$$ Brailey Sims (<https://www.mathshunter.edu.au/brailey/Research_papers/As%20Support%20Map%20Characterisation%20of%20the%20Opial%20Conditons.pdf>) characterised spaces with the strict Opial Property as those spaces such that whenever $x\_n$ converges weakly to $x\ne 0$, then $$\limsup\_{\substack{n\to\infty\\j x\_n\in J x\_n}}\langle x\,,j x\_n\rangle>0\,.$$
(The $\limsup$ may be equivalently replaced with $\liminf$). From this it is easy to deduce that the sequence spaces $\ell^p$, where $1\le p<\infty$ have Opial’s Property.


