[proofplan] We introduce the shifted operator $S:=N-\lambda I_H$ and prove directly that $S$ is normal. The equality of norms then follows by expanding $\|Sx\|_H^2$ using the adjoint, commuting $S^*S$ to $SS^*$ by normality, and recognizing the result as $\|S^*x\|_H^2$. Finally, substituting $S^*=N^*-\overline{\lambda}I_H$ transfers any bounded-below estimate from $S$ to $S^*$ with the same constant. [/proofplan]

[step:Show that the scalar shift of $N$ is normal]Fix $\lambda\in\mathbb C$, and define the [bounded linear operator](/page/Bounded%20Linear%20Operator) $S\in\mathcal L(H)$ by \begin{align*} S:=N-\lambda I_H. \end{align*} For every $x,y\in H$, \begin{align*} (Sx,y)_H=((N-\lambda I_H)x,y)_H=(Nx,y)_H-\lambda(x,y)_H. \end{align*} Using the definition of the [Hilbert space](/page/Hilbert%20Space) adjoint and the convention that $(\cdot,\cdot)_H$ is linear in the first argument, this becomes \begin{align*} (Sx,y)_H=(x,N^*y)_H-(x,\overline{\lambda}y)_H=(x,(N^*-\overline{\lambda}I_H)y)_H. \end{align*} Therefore \begin{align*} S^*=N^*-\overline{\lambda}I_H. \end{align*} Now compute \begin{align*} S^*S=(N^*-\overline{\lambda}I_H)(N-\lambda I_H) \end{align*} and hence \begin{align*} S^*S=N^*N-\lambda N^*-\overline{\lambda}N+|\lambda|^2I_H. \end{align*} Similarly, \begin{align*} SS^*=(N-\lambda I_H)(N^*-\overline{\lambda}I_H) \end{align*} and therefore \begin{align*} SS^*=NN^*-\overline{\lambda}N-\lambda N^*+|\lambda|^2I_H. \end{align*} Since $N$ is normal, $N^*N=NN^*$. Thus $S^*S=SS^*$, so $S$ is normal.[/step]

[guided]The point of introducing $S:=N-\lambda I_H$ is that the desired identity is exactly the standard norm identity for a normal operator and its adjoint, applied to this shifted operator. We first need to identify its adjoint. For $x,y\in H$, the operator $S$ satisfies \begin{align*} (Sx,y)_H=((N-\lambda I_H)x,y)_H=(Nx,y)_H-\lambda(x,y)_H. \end{align*} By the definition of $N^*$, we have $(Nx,y)_H=(x,N^*y)_H$. Because the Hilbert space [inner product](/page/Inner%20Product) is linear in the first argument and conjugate-linear in the second, the scalar term satisfies \begin{align*} \lambda(x,y)_H=(x,\overline{\lambda}y)_H. \end{align*} Therefore \begin{align*} (Sx,y)_H=(x,N^*y)_H-(x,\overline{\lambda}y)_H=(x,(N^*-\overline{\lambda}I_H)y)_H. \end{align*} Since this identity holds for all $x,y\in H$, the defining property of the adjoint gives \begin{align*} S^*=N^*-\overline{\lambda}I_H. \end{align*} We next verify normality of $S$ rather than assuming it. Multiplying the two factors gives \begin{align*} S^*S=(N^*-\overline{\lambda}I_H)(N-\lambda I_H). \end{align*} Expanding by bilinearity of operator composition and using $I_HN=N$ and $N^*I_H=N^*$ gives \begin{align*} S^*S=N^*N-\lambda N^*-\overline{\lambda}N+|\lambda|^2I_H. \end{align*} Likewise, \begin{align*} SS^*=(N-\lambda I_H)(N^*-\overline{\lambda}I_H) \end{align*} and expansion gives \begin{align*} SS^*=NN^*-\overline{\lambda}N-\lambda N^*+|\lambda|^2I_H. \end{align*} The only possible difference between these two expressions is the first term. Since $N$ is normal, $N^*N=NN^*$. Substituting this equality into the two expansions gives $S^*S=SS^*$, so the shifted operator $S=N-\lambda I_H$ is normal.[/guided]

[step:Convert normality of the shift into equality of the two norms] Let $x\in H$. Since $S$ is normal, $S^*S=SS^*$. Using the defining property of the adjoint, \begin{align*} \|Sx\|_H^2=(Sx,Sx)_H=(S^*Sx,x)_H. \end{align*} By normality of $S$, \begin{align*} (S^*Sx,x)_H=(SS^*x,x)_H. \end{align*} Applying the defining property of the adjoint again, now to $S$ and the vector $S^*x$, gives \begin{align*} (SS^*x,x)_H=(S^*x,S^*x)_H=\|S^*x\|_H^2. \end{align*} Thus \begin{align*} \|Sx\|_H^2=\|S^*x\|_H^2. \end{align*} Both sides are non-negative [real numbers](/page/Real%20Numbers), so taking square roots yields \begin{align*} \|Sx\|_H=\|S^*x\|_H. \end{align*} Substituting $S=N-\lambda I_H$ and $S^*=N^*-\overline{\lambda}I_H$ gives \begin{align*} \|(N-\lambda I_H)x\|_H=\|(N^*-\overline{\lambda}I_H)x\|_H. \end{align*} [/step]

[step:Transfer the bounded-below estimate to the adjoint shift] Assume that $N-\lambda I_H$ is bounded below with constant $c>0$, meaning that \begin{align*} \|(N-\lambda I_H)x\|_H\ge c\|x\|_H \end{align*} for every $x\in H$. By the norm equality already proved, for every $x\in H$, \begin{align*} \|(N^*-\overline{\lambda}I_H)x\|_H=\|(N-\lambda I_H)x\|_H. \end{align*} Combining the two displayed identities gives \begin{align*} \|(N^*-\overline{\lambda}I_H)x\|_H\ge c\|x\|_H \end{align*} for every $x\in H$. Since $(N-\lambda I_H)^*=N^*-\overline{\lambda}I_H$, this proves that $(N-\lambda I_H)^*$ is bounded below with the same constant $c$. [/step]