Let $V$ be an inner product space and $T:V \to V$ be a linear operator. Show that $N(T)^\perp=R(T^*)$.
Trial: $N(T)=R(T^*)^\perp$ since $y \in N(T) \iff Ty=0 \iff \langle x,Ty \rangle=0 \quad \forall x \iff \langle T^*x,y \rangle=0 \quad \forall x \iff y \in R(T^*)^\perp$. Taking $\perp$ both sides, $N(T)^\perp=(R(T^*)^\perp)^\perp$. I know that for finite dimensional $W$, $(W^\perp)^\perp=W$. But in this case it may not be finite dimensional. Does it need another approach?
Because $\mathcal{N}(T)^{\perp}$ is closed, and because $\mathcal{R}(T^{\star})$ may not be closed, this is not true. Pick $T=T^{\star}$ so that it has an inverse, but not a continuous inverse, and you'll have a counter-example to what you're trying to show. For example, let $V=l^{2}$ be the classical Hilbert space of square-summable sequences $\{ x_{n}\}_{n=1}^{\infty}$, and let $T\{ x_{n}\} = \{ \frac{1}{n}x_{n}\}$. Then $T^{\star}=T$. Furthermore, $T x=0$ iff $x=0$, but $T^{\star}$ is not onto because $\{ \frac{1}{n}\}$ is not in the range of $T=T^{\star}$. So $\mathcal{N}(T)^{\perp}=V$, but $\mathcal{R}(T)\ne V$.
