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.



Coercive Symmetric Bilinear form on a Hilbert space 
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 $v\_k=e\_k$, $k=1,2$.


