Prove Weak convergence in $H$.

Question

Let the sequence $(x_n)$ of points of the Hilbert space $H$ weakly coincide to the point $x ∈ H$. Prove that for an arbitrary point $y ∈ H-\{x\}$ has

the place is uneven

$\lim_{n→∞}||x_n − x|| < \lim_{n→∞}||x_n − y|| $.

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Answer 1

By replacing $x_n$ by $x_n-x$ and $y$ by $y-x$ we may assume that $x_n\to 0$ weakly and $y\neq 0.$ Then we need to show that $$\liminf\|x_n\|<\liminf \|x_n-y\|,\quad y\neq 0\quad (*)$$
We have $$\|x_n-y\|^2=\|x_n\|^2+\|y\|^2-2{\rm Re}\,\langle x_n,y\rangle$$ Taking $\liminf$ gives $$\liminf\|x_n-y\|^2=\liminf\|x_n\|^2+\|y\|^2$$ which implies $(*).$ Still we can get an explicit lower bound of the difference. Let $c=\sup\|x_n\|.$ Then $$\liminf\|x_n-y\|-\liminf\|x_n\|\\ \ge {\|y\|^2\over 2c+\|y\| }$$