If $H$ is a Hilbert space and $T$ is a bounded linear operator on $H$ with the following property
$$0 = \left(TH\right)^\perp = \ker T^*$$
where $T^*$ is injective, then $TH = H$.
Discussion
I've been looking at this for a few days now and I don't know why it's true. The only thing I can figure out is that $\left(TH\right)^\perp = 0 \implies TH$ is dense. But the role $T^*$ plays here is perplexing me. Is this true for infinite dimensional Hilbert spaces?
Don't think this is true.
Consider $$ T :l^2 (\mathbb N) \rightarrow l^2( \mathbb N )$$ $$ (x_n)_ n \mapsto \left (\frac {x_n}{n} \right )_n$$
Then $T$ is compact self adjoint and $ Range \ T \supset c_c(\mathbb N) $ . So in particular $ Range \ T ^ {\perp} =0= Ker \ T^* $ .
But $ T$ is not surjective as there's no preimage of $\left (\frac {1}{n}\right )_n$