Suppose that $H$ is an inner product space and let $T: H \longrightarrow$ H be a linear operator. Show that, if for every $x \in X$, there exists a $c>0$ such that
$$ \langle Tx , x \rangle \geq c\|x\|^2 $$ then $\text{Im}T$ is dense.
Discussion:
A short proof is using a theorem which states that $D \subset H$ is dense if $D^\perp = \{0\}$, which is obviously true in this case.
However, I seek for a more "analytic" approach, if there is one.
Thoughts:
Let $x \in H$. Then
$$ \lambda = \frac{\langle Tx , x \rangle}{\|x\|^2} $$
is the magnitude of the orthogonal projection of $Tx$ along $x$. Then, for every $x \in H$, there exists a $c > 0$ such that $$ \lambda \geq c $$
Also, applying the pythagorean theorem:
$$ \|Tx - \lambda x\|^2 = \|Tx\|^2 - \lambda \|x\|^2 \leq \|Tx\|^2 - c \|x\|^2 $$
Is this approach at all close to showing that, for every $\varepsilon > 0$
$$ \|Tx-x\| < \varepsilon $$