Limits in an inner product space

Question

Let $X$ be a real or complex inner product space, and $\{a_n\}$ a sequence in $X$ with $a\in X$.
I want to prove the following:

1. $\lim_{n\rightarrow \infty}(a_n,b)=(a,b)$ for all $b\in X$ and $\lim_{n\rightarrow \infty}||a_n||=||a||\implies\lim_{n\rightarrow \infty}a_n=a$
2. $\lim_{n\rightarrow \infty}a_n=a\iff\lim_{n\rightarrow \infty}(a_n,b)=(a,b)$ uniformly on the unit sphere $\{b\in X:||b||=1\}$.

What I know:
1 I thought $\left\| a - a_n \right\|^2 = (a-a_n,a-a_n) = \left\| a \right\|^2 + \left\| a_n \right\|^2 - 2 (a_n ,a) \to 2\left\| a \right\|^2 - 2 \left\| a \right\|^2 = 0,$ for $n\rightarrow \infty$. Is this correct?

2 I have no idea where to start this. Any hint to start me off would be greatly appreciated!

Answer 1

For the $\Rightarrow$ part:
We have $|(a_n,b)-(a,b)|=|(a_n-a,b)|$. The Cauchy Schwarz inequality gives us $$|(a_n-a,b)|\leq ||a_n-a||\cdot||b||=||a_n-a||\rightarrow0$$ as $n\rightarrow\infty$.

For the $\Leftarrow$ part:
We have $$||a_n-a||=\sup_{b\in X}\{|(a_n-a,b)|:||b||=1\}=\sup_{b\in X}\{|(a_n,b)-(a,b)|:||b||=1\}\rightarrow0$$ as $n\rightarrow\infty$.