Coercive Symmetric Bilinear form on a Hilbert space

Question

I need to show one of the two following equivalent results. If true, it must be a simple proof but I do not seem to be able to make it work. Thank you in advance.

1) Consider a continuous symmetric bilinear form $B$ on a Hilbert space. Let P be a closed subspace of that Hilbert space over which the form is coercive, i.e, there exists $\alpha\in \mathbb{R}$ such that $B(p,p)\geq \alpha \| p \|^2$ for all $p\in P$. Take the union of $P$ with a one dimensional subspace generated by an element which we will call $y$. Let say $B(y,y)>0$. Then then I need to show that $B$ is also coercive on the new subspace.

Another result that would do the trick for me goes as follows:

2) Consider a continuous symmetric bilinear $B$ form on a Hilbert space. If $v_n, {n\in \mathbb{Z}}$ is a basis of the Hilbert space and there exists a $\alpha$ such that $B(v_n,v_n)\geq \alpha \| v_n \|^2$ for all $v_n$, then $B$ is coercive, i,e, there exists $\alpha'\in \mathbb{R}$ such that $B(h,h)\geq \alpha' \| h \|^2$ for any $h$ in the Hilbert space.

Answer 1

This is not true. In $R^2$ consider $B(x,y) = x_1y_1 + x_2y_2 - x_1y_2 - y_1x_2$. Then $B(e_1,e_1) = B(e_2,e_2) = 1$ but $B(e_1+e_2,e_1+e_2)=0$. For 1) take $P=span(e_1)$, $y=e_2$, for 2) take $v_k=e_k$, $k=1,2$.