[proofplan] We prove the four kernel-range duality identities directly from the defining property $(Tx, y)_K = (x, T^*y)_H$ of the adjoint. Parts (i) and (ii) follow by chaining this identity with the non-degeneracy of the inner product. Parts (iii) and (iv) are obtained by taking orthogonal complements of both sides and applying the Hilbert space identity $(M^\perp)^\perp = \overline{M}$. [/proofplan] [step:Show that $\ker(T^*) = \operatorname{Range}(T)^\perp$ via the adjoint definition] Let $y \in K$. The following chain of equivalences holds: \begin{align*} y \in \ker(T^*) &\iff T^*y = 0 \\ &\iff (x, T^*y)_H = 0 \quad \text{for all } x \in H \\ &\iff (Tx, y)_K = 0 \quad \text{for all } x \in H \\ &\iff y \in \operatorname{Range}(T)^\perp. \end{align*} The second equivalence uses non-degeneracy of $(\cdot, \cdot)_H$: if $(x, v)_H = 0$ for every $x \in H$, then $v = 0$. The third equivalence is the defining property of the Hilbert adjoint: $(Tx, y)_K = (x, T^*y)_H$ for all $x \in H$. [guided] We want to show that the kernel of $T^*$ consists precisely of those vectors in $K$ that are orthogonal to every element of $\operatorname{Range}(T)$. The key tool is the defining property of the adjoint: $(Tx, y)_K = (x, T^*y)_H$ for all $x \in H$ and $y \in K$. Start from $y \in \ker(T^*)$, meaning $T^*y = 0$. Then for every $x \in H$: \begin{align*} (Tx, y)_K = (x, T^*y)_H = (x, 0)_H = 0. \end{align*} Since $x$ ranges over all of $H$, the element $Tx$ ranges over all of $\operatorname{Range}(T)$, so $y \perp \operatorname{Range}(T)$. Conversely, suppose $y \in \operatorname{Range}(T)^\perp$, so $(Tx, y)_K = 0$ for all $x \in H$. The adjoint property gives $(x, T^*y)_H = (Tx, y)_K = 0$ for all $x \in H$. By non-degeneracy of the inner product on $H$ -- if $(x, v)_H = 0$ for every $x \in H$, then choosing $x = v$ gives $\|v\|_H^2 = 0$, hence $v = 0$ -- we conclude $T^*y = 0$, i.e., $y \in \ker(T^*)$. [/guided] [/step] [step:Show that $\ker(T) = \operatorname{Range}(T^*)^\perp$ by the same argument with roles exchanged] Let $x \in H$. Then $x \in \ker(T)$ if and only if $Tx = 0$, if and only if $(Tx, y)_K = 0$ for all $y \in K$ (by non-degeneracy of $(\cdot, \cdot)_K$), if and only if $(x, T^*y)_H = 0$ for all $y \in K$ (by the adjoint property), if and only if $x \in \operatorname{Range}(T^*)^\perp$. [/step] [step:Derive the closure identities (iii) and (iv) from the double-complement identity] For any subspace $M$ of a [Hilbert space](/page/Hilbert%20Space), the identity $(M^\perp)^\perp = \overline{M}$ holds. Applying this to part (i): \begin{align*} \overline{\operatorname{Range}(T)} = \bigl(\operatorname{Range}(T)^\perp\bigr)^\perp = \bigl(\ker(T^*)\bigr)^\perp = \ker(T^*)^\perp. \end{align*} This is part (iii). Applying the same reasoning to part (ii): \begin{align*} \overline{\operatorname{Range}(T^*)} = \bigl(\operatorname{Range}(T^*)^\perp\bigr)^\perp = \bigl(\ker(T)\bigr)^\perp = \ker(T)^\perp. \end{align*} This is part (iv). [guided] Why does $(M^\perp)^\perp = \overline{M}$ hold? The inclusion $M \subseteq (M^\perp)^\perp$ is immediate: if $x \in M$, then $(x, y) = 0$ for all $y \in M^\perp$, so $x \in (M^\perp)^\perp$. Since $(M^\perp)^\perp$ is closed (as an orthogonal complement), $\overline{M} \subseteq (M^\perp)^\perp$. For the reverse inclusion, suppose $x \in (M^\perp)^\perp \setminus \overline{M}$. By the orthogonal decomposition $H = \overline{M} \oplus \overline{M}^\perp$, write $x = m + w$ with $m \in \overline{M}$ and $w \in \overline{M}^\perp = M^\perp$. Since $x \in (M^\perp)^\perp$, $(x, w)_H = 0$. But also $(m, w)_H = 0$ since $m \in \overline{M}$ and $w \in M^\perp$. So $\|w\|_H^2 = (w, w)_H = (x - m, w)_H = (x, w)_H - (m, w)_H = 0$, giving $w = 0$ and $x = m \in \overline{M}$. Applying this to part (i): $\operatorname{Range}(T)^\perp = \ker(T^*)$, so $(\ker(T^*))^\perp = (\operatorname{Range}(T)^\perp)^\perp = \overline{\operatorname{Range}(T)}$. The same identity applied to part (ii) yields $\overline{\operatorname{Range}(T^*)} = \ker(T)^\perp$. [/guided] [/step] [step:Conclude the density characterisation] Part (iii) states $\overline{\operatorname{Range}(T)} = \ker(T^*)^\perp$. The range $\operatorname{Range}(T)$ is dense in $K$ if and only if $\overline{\operatorname{Range}(T)} = K$, which holds if and only if $\ker(T^*)^\perp = K$, i.e., $\ker(T^*) = \{0\}$. Thus $\operatorname{Range}(T)$ is dense if and only if $T^*$ is injective. The analogous statement for $\operatorname{Range}(T^*)$ follows from part (iv) by the same reasoning. [/step]