# lower semi-bounded imply symmetric

A quadratic form is a map $$q: Q(q) \times Q(q) \rightarrow \mathbb{C}$$, where $$Q(q)$$ is a dense linear subset of the Hilbert space $$H$$. If $$q(\phi,\psi)=\overline{q(\psi,\phi)}$$, then we say q is symmetric. If $$q(\phi,\phi)\geq -M||\phi||^2$$ for some $$M$$, we say $$q$$ is semibounded.

If $$q$$ is semibounded, then it is automatically symmetric if $$H$$ is complex.

Anyone could give my a hint for the proof?

I have made an attempt to solve this: Let $$\phi,\psi \in H$$. We know that $$q(\phi+\psi,\phi+\psi)\in \mathbb{R}$$ and $$q(\phi+i\psi,\phi+i\psi)\in \mathbb{R}$$. $$q(\phi+\psi,\phi+\psi)=q(\phi,\phi)+q(\psi,\psi)+q(\psi,\phi)+q(\phi,\psi)(1)$$ $$q(\phi+i\psi,\phi+i\psi)=q(\phi,\phi)+q(\psi,\psi)-iq(\psi,\phi)+iq(\phi,\psi) \;(2)$$
From (2), we know that $$q(\psi,\phi)=q(\phi,\psi)$$. From (1), $$q(\psi,\phi)+q(\phi,\psi)$$ is real. Thus, $$q(\phi,\psi)$$ and $$q(\phi,\psi)$$ are real. $$q(\psi,\phi)=\overline{q(\phi,\psi)}$$