Let $(X, \langle \cdot, \cdot \rangle)$ be unitary space and let $(x_n)$ be a sequence in $X$.

Question

Let $(X, \langle \cdot, \cdot \rangle)$ be unitary space and let $(x_n)$ be a sequence in $X$. Suppose that $\Vert x_n \Vert \longrightarrow \Vert x_n \Vert$ and $\langle x_n, x_0 \rangle \longrightarrow \Vert x_0 \Vert ^2$, show that $x_n \longrightarrow x_0.$

My idea was to check the convergence of $\Vert x_n - x_0 \Vert ^2$.

So $\Vert x_n - x_0 \Vert ^2 = \langle x_n-x_0, x_n-x_0 \rangle = \Vert x_n \Vert^2 - 2\langle x_n, x_0\rangle + \Vert x_0 \Vert^2 \longrightarrow 0 $ by these two assumptions at the beginning. Since inner product is $0$ only for zero vector is it possible to deduce that $x_n \longrightarrow x_0$?

Answer 1

Almost there. $\langle x_n,x_0\rangle \to \|x_0\|^2 \implies \lim_{n \to \infty} \|x_0\|^2+\|x_n\|^2-2\langle x_n, x_o \rangle=?$

In particular, notice that $\|x_n\| \to x_0$ and limits distribute over addition.

Answer 2

No. You have missed the fact that $\langle x_n, x_0 \rangle \to ||x_0||^2$. Use it. Then the right hand side converges to $0$, which is what you need.