Prove $T$ is surjective if $(Tx,x)\geq k(x,x)$ in Hilbert space Suppose $H$ is a Hilbert space, and $T$ is a continuous operator satisfies 
$$(Tx,x)\geq k(x,x), k>0,x\in H$$
How to prove that $T$ is onto? 
 A: You can do it in the following steps:


*

*Prove that $A$ is injective.

*Prove that $A^{-1} : R(A) \to H$ is bounded (where $R(A)$ denotes the range of $A$). In fact, it is easy to see that $\|A^{-1}\| \leq \frac{1}{k}$.

*Prove that $R(A)$ is complete : Take a Cauchy sequence $y_n = A(x_n)$, and then show that $(x_n)$ is Cauchy. Now appeal to the completeness of $H$ and the continuity of $A$.

*Prove that $R(A)^{\perp} = \{0\}$: This is easy because if $y\in R(A)^{\perp}$, then $\langle A(y),y\rangle = 0$

*Conclude that $R(A) = H$, so $A$ is surjective.
A: The inequality implies that  $T $ is positive. In particular it is selfadjoint and so its spectrum is contained in the numerical range. It follows that $\sigma (T)\subset [k,\|T\|] $, so $0\not\in\sigma (T) $; thus, $T $ is invertible. 
A: I'm assuming your $T$ is linear. If $(Tx,x)\ge k(x,x)$ for some fixed $k > 0$ and for all $x$, then $T$ is injective because $Tx=0$ implies $x=0$. The range of $T$ is dense because, if $x \perp \overline{\mathcal{R}(T)}$, then $0=(Tx,x)\ge k(x,x)$, which again implies $x=0$. The inverse is bounded on its domain $\mathcal{R}(T)$ because
$$
                  k\|x\|^2 = (Tx,x) \\
                  k\|T^{-1}y\|^2 = (y,T^{-1}y) \le \|y\|\|T^{-1}y\|,\;\; y\in\mathcal{R}(T), \\
                 \implies k\|T^{-1}y\| \le \|y\|,\;\; y\in\mathcal{R}(T).
$$
So $T$ is a continuous bijection between $H$ and $\mathcal{R}(T)$, which forces $\mathcal{R}(T)$ to be closed. So $\mathcal{R}(T)=\overline{\mathcal{R}(T)}=H$. (You can see this directly by letting $y\in H$ be given, choosing $\{y_n \} \in \mathcal{R}(T)$ that converges to $y$, finding the unique sequence $\{ x_n \}$ so that $y_n = Tx_n$, and concluding that $\{ x_n \}$ is Cauchy because $T^{-1}$ is bounded. Hence $\{ x_n \}$ converges to some $x_0$, which forces $Tx_0 = \lim_n Tx_n = \lim_n y_n = y$.)
