[proofplan]
Fix a complex number $\lambda$ and study the shifted operator $T=N-\lambda I_H$. The key point is that $T$ is again normal, and for a normal operator the kernels of $T$ and $T^*$ agree because $\|Tx\|_H=\|T^*x\|_H$ for every $x\in H$. If $T$ is injective, then $\ker T^*=\{0\}$, while the orthogonal complement of $\operatorname{Range}(T)$ is exactly $\ker T^*$. Therefore the range of every injective shift is dense, which rules out residual spectral points.
[/proofplan]
[step:Reduce the residual-spectrum assertion to density of injective shifts]
Let $\lambda\in\mathbb C$ be arbitrary, and let $I_H\in\mathcal L(H)$ denote the identity operator on $H$. Define
\begin{align*}
T:=N-\lambda I_H\in\mathcal L(H).
\end{align*}
By the definition of residual spectrum, $\lambda\in\sigma_r(N)$ exactly when $T$ is injective and $\operatorname{Range}(T)$ is not dense in $H$. It is therefore enough to prove that whenever $T$ is injective, one has
\begin{align*}
\overline{\operatorname{Range}(T)}=H.
\end{align*}
[/step]
[step:Show that the shifted operator is normal]
The adjoint of $T=N-\lambda I_H$ is
\begin{align*}
T^*=N^*-\overline{\lambda}I_H.
\end{align*}
Using $NN^*=N^*N$, the commutation of $I_H$ with every operator in $\mathcal L(H)$, and scalar associativity, we compute
\begin{align*}
T^*T=(N^*-\overline{\lambda}I_H)(N-\lambda I_H).
\end{align*}
Expanding the product gives
\begin{align*}
T^*T=N^*N-\lambda N^*-\overline{\lambda}N+|\lambda|^2I_H.
\end{align*}
Similarly,
\begin{align*}
TT^*=(N-\lambda I_H)(N^*-\overline{\lambda}I_H)
\end{align*}
and hence
\begin{align*}
TT^*=NN^*-\overline{\lambda}N-\lambda N^*+|\lambda|^2I_H.
\end{align*}
Since $NN^*=N^*N$, these two expressions are equal. Thus
\begin{align*}
T^*T=TT^*,
\end{align*}
so $T$ is normal.
[/step]
[step:Use normality to identify the kernels of $T$ and $T^*$]
For every $x\in H$, the Hilbert-space norm satisfies
\begin{align*}
\|Tx\|_H^2=(Tx,Tx)_H.
\end{align*}
By the defining property of the adjoint,
\begin{align*}
(Tx,Tx)_H=(T^*Tx,x)_H.
\end{align*}
Since $T$ is normal, $T^*T=TT^*$, so
\begin{align*}
(T^*Tx,x)_H=(TT^*x,x)_H.
\end{align*}
Again using the defining property of the adjoint,
\begin{align*}
(TT^*x,x)_H=(T^*x,T^*x)_H=\|T^*x\|_H^2.
\end{align*}
Therefore
\begin{align*}
\|Tx\|_H=\|T^*x\|_H
\end{align*}
for every $x\in H$. It follows that $Tx=0$ if and only if $T^*x=0$, and hence
\begin{align*}
\ker T=\ker T^*.
\end{align*}
[guided]
We want to connect injectivity of $T$ with information about the range of $T$. The bridge is the adjoint, so first we prove that $T$ and $T^*$ have the same kernel. The normality of $T$ is exactly what allows this.
Fix $x\in H$. By definition of the Hilbert-space norm,
\begin{align*}
\|Tx\|_H^2=(Tx,Tx)_H.
\end{align*}
The adjoint $T^*$ is defined by the identity $(Tu,v)_H=(u,T^*v)_H$ for all $u,v\in H$. Applying this with $u=x$ and $v=Tx$ gives
\begin{align*}
(Tx,Tx)_H=(x,T^*Tx)_H.
\end{align*}
Because the [inner product](/page/Inner%20Product) is conjugate-symmetric, this equals $(T^*Tx,x)_H$ as a real non-negative number. Since $T$ is normal, $T^*T=TT^*$, and therefore
\begin{align*}
(T^*Tx,x)_H=(TT^*x,x)_H.
\end{align*}
Applying the adjoint identity once more, now to $T$ and the vector $T^*x$, gives
\begin{align*}
(TT^*x,x)_H=(T^*x,T^*x)_H=\|T^*x\|_H^2.
\end{align*}
Combining the displayed equalities yields
\begin{align*}
\|Tx\|_H^2=\|T^*x\|_H^2.
\end{align*}
Both sides are non-negative [real numbers](/page/Real%20Numbers), so
\begin{align*}
\|Tx\|_H=\|T^*x\|_H.
\end{align*}
Thus $Tx=0$ exactly when $\|Tx\|_H=0$, exactly when $\|T^*x\|_H=0$, exactly when $T^*x=0$. Hence
\begin{align*}
\ker T=\ker T^*.
\end{align*}
[/guided]
[/step]
[step:Convert injectivity into density of the range]
Assume that $T$ is injective. Then
\begin{align*}
\ker T=\{0\}.
\end{align*}
From the preceding step,
\begin{align*}
\ker T^*=\{0\}.
\end{align*}
We claim that
\begin{align*}
\operatorname{Range}(T)^\perp=\ker T^*.
\end{align*}
Indeed, for $y\in H$, the condition $y\in\operatorname{Range}(T)^\perp$ means that
\begin{align*}
(Tx,y)_H=0
\end{align*}
for every $x\in H$. By the defining property of the adjoint, this is equivalent to
\begin{align*}
(x,T^*y)_H=0
\end{align*}
for every $x\in H$. Taking $x=T^*y$ gives $\|T^*y\|_H^2=0$, so $T^*y=0$. Conversely, if $T^*y=0$, then $(Tx,y)_H=(x,T^*y)_H=0$ for every $x\in H$, so $y\in\operatorname{Range}(T)^\perp$. Thus the claimed identity holds.
Therefore
\begin{align*}
\operatorname{Range}(T)^\perp=\{0\}.
\end{align*}
For any subspace $M\subset H$, $\overline{M}=H$ if and only if $M^\perp=\{0\}$. Applying this with $M=\operatorname{Range}(T)$ gives
\begin{align*}
\overline{\operatorname{Range}(T)}=H.
\end{align*}
[/step]
[step:Exclude every possible residual spectral point]
We have shown that for an arbitrary $\lambda\in\mathbb C$, if $N-\lambda I_H$ is injective, then $\operatorname{Range}(N-\lambda I_H)$ is dense in $H$. Hence no $\lambda\in\mathbb C$ can satisfy the defining condition for membership in the residual spectrum of $N$. Therefore
\begin{align*}
\sigma_r(N)=\varnothing.
\end{align*}
[/step]
