# Unit sphere weakly dense in unit ball

I'm studying for an exam and came across a problem: I want to prove that the unit sphere in a Hilbert space $\mathcal{H}$ is weakly dense in the unit ball.

I already had to prove that the unit ball is weakly closed, so one direction is done. What remains to be seen is that the unit ball is contained in the weak closure of the sphere. I suspect that orthonormal bases will need to be used here; the problem also had us prove that every orthonormal sequence in $\mathcal{H}$ converges weakly to $0$ which I did.

There's another similar question here: Prove: The weak closure of the unit sphere is the unit ball. but that deals with normed spaces as opposed to Hilbert ones and is beyond the scope of my course from the looks of it.

• Hilbert spaces are normed spaces, via $||x|| = \sqrt(<x,x>)$ – Mark Joshi Dec 9 '14 at 5:17
• @MarkJoshi Yes, but Hilbert spaces have the additional bonus of being complete wrt the norm, and the results in the section were proved in the context of Hilbert spaces. – Lost Dec 9 '14 at 6:13

A basic neighbourhood of a point $p$ in the weak topology can be written in the form $$U(y_1, \ldots, y_n; p) = \{x: \left|\langle y_j, x - p \rangle\right| < 1, j = 1 \ldots n\}$$ where $\{y_1, \ldots, y_n\}$ is any finite set of vectors in $\mathcal H$. If $\|p \| < 1$, find $v$ so $\langle y_j, v\rangle = 0$ for $j = 1\ldots n$ and add an appropriate multiple of $v$ to $p$ to get a unit vector in $U(y_1, \ldots, y_n; p)$.
• Great remark! This proof also works for infinite-dimensional normed vector spaces :) By taking the $y_j\in X^*$. – Juan Carlos Ortiz Feb 5 '20 at 20:34