In the following, we suppose that $H_1$ and $H_2$ are Hilbert spaces, and let $F:H_1\mapsto H_2$ be a linear continuous operator, and its adjoint $F^*$ is defined by \[ (1)\hspace{20pt}\inner{ F^*(x_2) }{ x_1 }=\inner{ x_2 }{ F(x_1) },\quad\forall\,x_1\in H_1,\;x_2\in H_2. \] Then the following statements hold #$F(H_1)$ is dense in $H_2$ if and only if $F^*$ is injective. #$F$ is onto if and only if $\norm{F^*(x_2)}\geq c\norm{x_2}$ holds for all $x_2\in H_2$ and some $c>0$. //Proof of Statement $1$//: Given that $F(H_1)$ is dense in $H_2$, and if $F^*(x_2)=0$ for some $x_2\in H_2$, then by $(1)$, we see that \[ (2)\hspace{20pt}\inner{ x_2 }{ F(x_1) } =0,\quad\forall\,x_1\in H_1, \] which implies that $x_2=0$ by the density of $F(H_1)$. Conversely, given that $F^*$ is injective, if the closure of $F(H_1)$ is not the whole space $H_2$, then there exists some nonzero element $x_2\in H_2$ such that $(2)$ holds, which, together with $(1)$, shows that \[ \inner{ F^*(x_2) }{ x_1 } = 0,\quad\forall\,x_1\in H_1. \] Hence $F^*(x_2)=0$, which contradicts with that $F^*$ is injective. //Proof of Statement $2$//: Without loss of generosity, we can assume that $F$ is injective, otherwise, we can consider the induced map $\widetilde F : H_1/\text{ker}F\mapsto H_2$. Now since $F$ is injective and onto, we thus have $F$ is bijective and a isomorphism from $H_1$ to $H_2$. Then $F^{-1}:H_2\mapsto H_1$ is also bounded. Hence choosing $x_1=F^{-1}(x_2)$ in $(1)$ yields: \[ \inner{ F^*(x_2) }{ F^{-1}(x_2) }=\inner{ x_2 }{ x_2 }, \] which implies \[ \norm{F^*(x_2)}\geq \frac{1}{\norm{F^{-1}}} \norm{x_2}. \] We thus complete the proof. **Remark 1** If $F$ is a bijective linear continuous operator from $H_1$ to $H_2$, then $F^*$ is also a bijective linear continuous operator from $H_2$ to $H_1$, and we have \[ (F^{-1})^* = (F^*)^{-1}. \] **Remark 2** The results can be extended to the reflexive Banach spaces.