(S,(f,g)) is an inner product space, and $(v_n)_{n\in{\mathbb N}}$ is an orthonormal set of vectors.
a) Show that, though ($v_n$) is a bounded sequence in norm, there is no subsequence which converges.
b) How do I go about showing that $lim_{n\to{\infty}}$ 1/N $\sum$ $u_n$=0 in the norm?
Do I have to go about using the idea that this type of space is not compact? closed but not bounded? or bounded but not closed? I am having difficulties, pleas give details.
Part (a) is straightforward. If $v_n$'s are orthonormal, then $\|v_n\| = 1$ for all $n$. So it is a bounded sequence. For each $n \neq m$, $\| v_n - v_m \|^2 = \langle v_n - v_m, v_n - v_m \rangle = 2$. So it violates Cauchy's criterion for convergence.
The notation in part (b) seems confusing. What is $u_n$, should that be $v_n$? What is the range of the summation, from $1$ to $N$? My guess would be $\lim_{N \to \infty} \frac{\sum_{n=1}^N v_n}{N}$. Then
$$\| \frac{1}{N}\sum_{n=1}^N v_n \|^2 = \langle \frac{1}{N}\sum_{n=1}^N v_n, \frac{1}{N}\sum_{n=1}^N v_n \rangle = \frac{1}{N^2} \sum_{n=1}^N \|v_n\|^2 = \frac{1}{N} \to 0,$$
where the second equality follows from the linearity of inner product and orthogonality of $v_n$'s.