# In a Hilbert Space, if $\langle x,x_n \rangle \to 0$ then $\sup\{\|x_n\|:n=1,2,3,...\}<\infty$

Let $\mathbb{H}$ be a Hilbert space. Let $\{x_n\}$ be a sequence in $\mathbb{H}$ with the property that $\langle x,x_n \rangle\to 0$ as $n\to\infty$ for $x\in\mathbb{H}$. Show that $\sup\{\|x_n\|:n=1,2,3,...\}<\infty$.

So this is what I have:

As $\langle x,x_n \rangle\to 0$, then for $\epsilon > 0$ there exists $N\in\mathbb{N}$ such that $\|\langle x,x_n \rangle-0\|<\epsilon, \forall n\geq N$. Also, as $\mathbb{H}$ is a Hilbert space, then we know that Then either $x=0$ or $x_n\to 0$. If $x_n\to 0$ then $x_n$ is bounded by some value $b$, and therefore $\sup\{\|x_n\|:n=1,2,3,...\}\leq b\lt\infty$. Now if $x=0$ and $x_n\to y$ where $y\neq\infty$, then once again $x_n$ is bounded by some $b$ and $\sup\{\|x_n\|:n=1,2,3,...\}\leq b\lt \infty$. Finally, let $\langle x,x_n \rangle \to 0$, $x=0$ and $x_n\to\infty$, then $\lim_{n\to\infty}\langle x,x_n\rangle=\langle \lim_{n\to\infty} x,\lim_{n\to\infty}x_n\rangle=\langle x,\infty\rangle$ which is undefined (so this cannot be, based on the asumptions). Therefore for all valid cases, $\sup\{\|x_n\|:n=1,2,3,...\}\lt\infty$.

• You cannot conclude directly that either $x=0$ or $x_n\to 0$. You need to apply Uniform Boundedness principle here. Oct 20 '14 at 6:13
Let $T_nx = \langle x_n , x \rangle$. We have $\sup_n |T_n x| < \infty$, hence by the Banach Steinhaus theorem we have $\sup_n \|T_n\| < \infty$. Since $\|T_n\| = \|x_n\|$, we have the desired result.
• Cauchy Schwartz gives $\|T_n x\| \le \|x_n\| \|x\|$, so $\|T_n \| \le \|x_n \|$. Then $T_n x_n = \|x_n\|^2$, which gives equality. Oct 20 '14 at 20:35