# Concluding that a linear operator on a Hilbert space is invertible

Setting:

Let $H$ be a Hilbert space with two inner products, $\langle \cdot,\cdot\rangle$ and $[\cdot, \cdot]$, and $S:H\to H$ be a bounded linear operator such that for all $x,y\in H$, we have $\langle Sx,y \rangle = [x,y]$. Suppose further that the two inner products induce norms which are equivalent.

Under these hypotheses, can we conclude that $S$ is a bijection with bounded inverse?

• Hellinger-Toeplitz theorem implies that $S$ is bounded, Lax-Milgram thm then implies that $S$ is bijective (+ bounded & bijective implies that the inverse in bounded, say by open mapping theorem) – user8268 Jun 9 '14 at 19:16

We get the lower bound from $$\|Sx\|\,\|x\|\ge \langle Sx,x\rangle = [x,x] \ge c\langle x,x\rangle = c\|x\|^2 \tag1$$ where $\|\cdot \|^2 = \langle \cdot,\cdot\rangle$. Hence, the range of $S$ is closed. If $S$ is not surjective, there is a nonzero $y$ such that $\langle Sx,y\rangle =0$ for all $x$. But then $\langle Sy,y\rangle =0$, which implies $y=0$ by (1).