Show that $y \perp x_n$ and $x_n \rightarrow x$ together imply $x \perp y$

Question

I was looking at this problem here, but I am still confused.

Show that $y \perp x_n$ and $x_n \rightarrow x$ together imply $x \perp y$.

Is it enough to just say that $$<x_n,y>=0$$ then since $x_n \rightarrow x$ we have that $<x,y>=0$?

I had this other proof: Show that for a sequence $(x_n)$ in an inner product space the conditions $||x_n||\rightarrow ||x||$ and $<x_n,x>\rightarrow <x,x>$ imply convergence $x_n\rightarrow x$.

The answer to this one in my book was $$||x_n-x||^2=<x_n-x,x_n-x>$$ $$=<x_n,x_n>-<x_n,x>-<x,x_n>+<x,x>$$ $$=2||x||^2-2<x,x>$$ $$=0$$

So it seems like we just replaced $x_n$ with $x$ and $<x_n,x>$ with $<x,x>$. Is this always allowed like this?

Thanks.

Answer 1

The property being used here to "replace $x_n$ with $x$" is the continuity of inner product in each of its arguments. More precisely, fixing $y$, the following function $$ f(x)=\langle x,y\rangle $$ is continuous.

Then by continuity of $f$, we have $$ \lim_{n\to \infty}f(x_n)=f(x)\implies\lim_{n\to \infty}\langle x_n,y\rangle =\langle x,y\rangle $$ but the sequence is constant $0$, so you may conclude $\langle x,y\rangle=0$.

Answer 2

We define $f:H \to \mathbb{R}$ such that $\left\langle {f,x} \right\rangle = \left\langle {x,y} \right\rangle $. Easily to see $f \in {H^\prime }$. If we have $\mathop {\lim }\limits_{n \to + \infty } {x_n} = x$ we will have $$\mathop {\lim }\limits_{n \to + \infty } \left\langle {f,{x_n}} \right\rangle = \left\langle {f,x} \right\rangle .$$