= Solution
Use an <inner product> linear in the first variable. For each $y$, the bounded linear functional $x\mapsto\langle Tx,y\rangle$ is represented uniquely by a vector $T^*y$ under the <Riesz representation theorem>, so $\langle Tx,y\rangle=\langle x,T^*y\rangle$. Uniqueness and the conjugate-linearity in the second argument imply $T^*(ay+bz)=aT^*y+bT^*z$. Furthermore
$$
\|T^*y\|=\sup_{\|x\|=1}|\langle Tx,y\rangle|\leq\|T\|\|y\|.
$$
Thus the <adjoint operator> exists, is linear and bounded, and is unique because a vector orthogonal to every $x$ is zero.
A <normal operator> satisfies $T^*T=TT^*$. The unilateral <unilateral shift operator> $S(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$ on $\ell^2$ is bounded but not normal: $S^*S=I$, whereas $SS^*=I-P_{e_1}$. For general $T$,
$$
\|Tx\|^2-\|T^*x\|^2=\langle(T^*T-TT^*)x,x\rangle.
$$
Normality implies equality of norms. Conversely equality for every $x$ makes this self-adjoint operator's quadratic form zero; the <polarization identity> then gives $T^*T-TT^*=0$.
An <approximate eigenvalue> $\lambda$ has unit vectors $x_n$ with $\|(T-\lambda I)x_n\|\to0$. 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, $\|Ax\|\geq c\|x\|$. It is injective and has closed range: a convergent sequence $Ax_n$ forces $x_n$ to be Cauchy. The operator $A$ is normal, so $\|A^*x\|=\|Ax\|$ and $\ker A^*=0$. The identity $(\operatorname{ran}A)^\perp=\ker A^*$ 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 $\boxed{\sigma(T)=\sigma_{\rm ap}(T)}$.
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