Decomposing bilinear forms in Hilbert spaces

2013-01-12T08:02:35Z

You are given a complex Hilbert space $H$ with two equivalent Hilbert space structures $<,>$ and $<,>'$. Define $<,>''=<,> + <,>'$ to be the sum of our two scalar products. Suppose you are given a complex symmetric bilinear form $B: H\times H \to \Bbb C$, with norm $c$ over $<,>''$ i.e. $$B(x,x) \leq c(|x|^2 + |x|'^2) , \ x \in H. $$ Does there exist complex symmetric bilinear forms $A$ and $C$, such that $B=A+C$ and $$A(x,x) \leq c|x|^2 , \ x \in H, $$ $$C(x,x) \leq c|x|'^2 , \ x \in H. $$ It is clear that if $A$ and $C$ have norm less then $c$ with respect to the respective norms, then $B$ has norm less then $c$ with respect to the "sum" norm. My questions is, do all operatods $B$ arise this way?

Probably not. However I keep thinking about this on and off, and as embarassing as it is, I can't come up with a counterexample even in the $\Bbb C^n$ case.

Answer by Nik Weaver for Decomposing bilinear forms in Hilbert spaces

2013-01-12T11:33:52Z

I think you mean $|B(x,x)| \leq c(|x|^2 + |x|'^2)$, etc. As the problem is stated, you can take $A(x,y) = \langle x,y\rangle$ and $C(x,y) = B(x,y) - \langle x,y\rangle$.

Then the answer is no, there's a counterexample where $H$ is two-dimensional. First, let us not demand that $\langle\cdot,\cdot\rangle$ and $\langle\cdot,\cdot\rangle'$ be equivalent. Set $\langle \vec{x},\vec{y}\rangle = x_1\bar{y}_1$ and $\langle \vec{x},\vec{y}\rangle' = x_2\bar{y}_2$. Then $\langle\cdot,\cdot\rangle''$ is the standard inner product on ${\bf C}^2$ and the form $B(\vec{x},\vec{y}) = \frac{1}{2}(x_1 + x_2)(y_1 + y_2)$ has norm 1. But $|A(\vec{x},\vec{x})| \leq |\vec{x}|^2 = |x_1|^2$ for all $\vec{x}$ implies that $A(e_1,e_2) = A(e_2,e_1) = 0$, and similarly for $C$, so their sum cannot have nonzero off-diagonal terms. Thus, we cannot write $B = A + C$ for such $A$ and $C$.

Now modify that example by letting $\langle \vec{x},\vec{y}\rangle = (1-\epsilon)x_1\bar{y}_1 + \epsilon x_2 \bar{y}_2$ and $\langle \vec{x},\vec{y}\rangle' = \epsilon x_1\bar{y}_1 + (1 - \epsilon)x_2 \bar{y}_2$. For any $\epsilon > 0$, these are equivalent inner products. Let $B$ be as above; since $\langle\cdot,\cdot\rangle + \langle\cdot,\cdot\rangle'$ is the standard inner product for any $\epsilon$, $B$ has norm 1 independent of $\epsilon$. If you could find $A$ and $C$ as desired for any $\epsilon > 0$, then by compactness (take a cluster point) you could find $A$ and $C$ as desired for $\epsilon = 0$, which we just saw is impossible. So for some $\epsilon > 0$ the given $B$ cannot be decomposed in the desired way.

It's not hard to give explicit estimates on the off-diagonal entries of $A$ and $C$ in terms of $\epsilon$ (I guess $\sqrt{2\epsilon}$ is a bound), and this would allow you to give an explicit counterexample. But I prefer the preceding proof.

Decomposing bilinear forms in Hilbert spaces - MathOverflow

Decomposing bilinear forms in Hilbert spaces

Answer by Nik Weaver for Decomposing bilinear forms in Hilbert spaces