Subtracting the weak limit reduces the norm in the limit

Question

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 $\|x_n - \mathbf{1}_{[-2,-1]}\|_{\infty} = 1 = \|x_n\|_\infty$.

Answer 1

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.