 Solution

Use an inner product linear in the first variable. For each $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ is represented uniquely by a vector $T^{\ast }y$ under the Riesz representation theorem, so $⟨Tx,y⟩=⟨x,T^{\ast }y⟩$. Uniqueness and the conjugate-linearity in the second argument imply $T^{\ast }(ay+bz)=a{T}^{\ast }y+b{T}^{\ast }z$. FurthermoreThus the adjoint operator exists, is linear and bounded, and is unique because a vector orthogonal to every $x$ is zero.

A normal operator satisfies $T^{\ast }T=T{T}^{\ast }$. The unilateral unilateral shift operator $S(x_1,x_2,\dots )=(0,x_1,x_2,\dots )$ on ${\ell }^2$ is bounded but not normal: $S^{\ast }S=I$, whereas $S{S}^{\ast }=I-P_{e_1}$. For general $T$,Normality implies equality of norms. Conversely equality for every $x$ makes this self-adjoint operator's quadratic form zero; the polarization identity then gives $T^{\ast }T-T{T}^{\ast }=0$.

An approximate eigenvalue $\lambda$ has unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$. It is necessarily in the spectrum, since a bounded inverse would prevent this. Conversely suppose $\lambda$ is not an approximate eigenvalue. Then $A=T-\lambda I$ is bounded below, $\Vert Ax\Vert \ge c\Vert x\Vert$. It is injective and has closed range: a convergent sequence $A{x}_n$ forces $x_n$ to be Cauchy. The operator $A$ is normal, so $\Vert A^{\ast }x\Vert =\Vert Ax\Vert$ and $\ker A^{\ast }=0$. The identity $(\mathrm{ran}A{)}^{\perp }=\ker A^{\ast }$ makes its range dense. Closed and dense means surjective, and its inverse is bounded by $1/c$. Hence $\lambda$ is in the resolvent. This proves $\boxed{\sigma (T)={\sigma }_{\mathrm{ap}}(T)}$.