Extending bilinear form from subspace to whole space

Question

Let $X$ be a linear subspace of a Hilbert space $Y$. Let $a(\cdot,\cdot):X \times X \to \mathbb{R}$ be bilinear. Suppose I know what $a$ is on $X$.

Is there some theorem or other that tells me that $a$ can be uniquely extended to all of $Y$? Something similar to a bounded linear transform theorem? ($X$ is not necessarily dense in $Y$.)

are there other assumptions I can use?

Answer 1

Consider unique continuous extenstion $\overline{a}$ of $a$ from $X\times X$ to $\overline{X}\times \overline{X}$. It is always possible. Now we have bilinear form defined on Hilbert space $\overline{X}$. By $P$ we denote orthogonal projection from $Y$ to $\overline{X}$. Recall that there exist linear isometric bijection between bounded operators and bilinear forms $$ I_H:\mathcal{B}(H)\to\mathcal{Bil}(H,H) : T\mapsto((x,y)\mapsto \langle T(x),y\rangle) $$ where $H$ - is a Hilbert space.

The desired bilinear extension $b$ is $I_{Y}(I_{\overline{X}}^{-1}(\overline{a})P)$.

Answer 2

It seems that $a$ can always be extended to $Y$. Moreover, the extension is unique iff $X$ is dense in $Y$. It seems that these can be proved by a following way. You can extend $a$ onto the closure $\overline X$ of $X$ using a standard technique, using the limits of Cauchy sequences or filters.

Now consider the orthogonal complement $X’=\{y\in Y:(y,x)=0$ for each $x\in \overline X\}$ of $X$. Composing $a$ with the square $p\times p$ of the projection $p:Y\to \overline X$ parallel to $X’$, you should obtain an extension of $a$ onto $Y$.

The standard proofs suggest that the extension is unique iff there is no non-zero bilinear map $b$ on $Y$ such that the restriction of $b$ on $X$ is zero, which holds iff $X$ is dense in $Y$. I expect that the following ideas are keys to the proofs.

1. It is well known (see, for instance, Theorem 1.5.4 from Engelking’s “General topology”) and is easy to check that two maps to a Hausdorff space, which are coinciding on a dense set, are equal.

2. If $X$ is not dense in $Y$ then there should be a notrivial vector $y$ in $X’$, which may be used for a construction of an extension $b$ of $a$ onto $Y$, such that $b(y,y)=1$.