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For $n\le x$, write $F(n)=F(x)-{\int }_n^x{F}^{\prime }(t)dt$, multiply by $f(n)$ and sum. Interchanging a finite sum with the integral gives the Abel summation formulaApply it with $F(t)=t^{-s}$ and $S(t)={\sum }_{n\le t}{a}_n$. If $\sigma >\alpha$, the boundary term is $O(x^{\alpha -\sigma })\to 0$ and ${\int }_1^{\infty }|S(t)t^{-s-1}|dt$ converges. Therefore the Dirichlet series converges andChoose $C$ with $|S(t)|\le C{t}^{\alpha }$ for $t\ge 1$. For $\sigma >\alpha +\delta$,The Weierstrass M-test proves uniform convergence of $\sum I_m$ on this entire open half-plane. For fixed $s$, the same estimate with its actual $\sigma$ gives $I_m(s)=O(m^{\alpha -1-\sigma })$. Each $I_m$ is an entire function, since its integral has bounded positive $t$ range and can be differentiated in $s$ under an integrable bound on each compact set. Thus the locally uniform sum is holomorphic, and so is its product with $s$. This gives holomorphy from polynomial growth of arithmetic partial sums and proves holomorphy of $A$ for $\mathrm{Re}s>\alpha$. Multiplication by the unbounded factor $s$ does not assert uniform convergence of $A$ on a whole unbounded half-plane; local uniform convergence is what the holomorphy argument needs.

With $a_n=\mu (n)$ and the hypothetical bound, $A_{\mu }(s)$ is holomorphic for $\mathrm{Re}s>3/4$. On $\mathrm{Re}s>1$, the Euler product gives $\zeta (s)A_{\mu }(s)=1$. The identity theorem extends this as a meromorphic identity throughout the connected half-plane $\mathrm{Re}s>3/4$. Hence $\zeta$ has no zeros in that half-plane: at any zero, the product with the holomorphic $A_{\mu }$ would be zero. The pole at one is not a zero; $A_{\mu }$ has a compensating zero there.

If the real part vanishes identically, the norm-difference identity becomes equality, soThus the same substitution as above gives $\Vert Sz\Vert =\Vert z\Vert$ for every $z$. To prove surjectivity as well, observe that both $T$ and $-T$ satisfy the nonnegative-real-part hypothesis. The preceding part makes both $I+T$ and $I-T$ invertible. Their product $S=(I-T)(I+T{)}^{-1}$ is therefore invertible and onto. It is a surjective norm-preserving linear operator, so the equivalences in (a) prove

The four claims below show that $I+T$ has a bounded inverse and that the Cayley transform of a bounded accretive operator is contractive. For an arbitrary $z\in X$, write $z=(I+T)x$. ThenTaking the supremum over the unit ball gives

Use the convention that the inner product is linear in its first argument. The precise polarization identity isIf (1) holds, apply this formula before and after $U$ to get preservation of all inner products, proving (2). If (2) holds, the identity $⟨Ux,Uy⟩=⟨x,U^{\ast }Uy⟩$ gives $U^{\ast }U=I$. For $z=Ux$, surjectivity then gives $U{U}^{\ast }z=U{U}^{\ast }Ux=Ux=z$, proving (3).

From (3), $\Vert Ux{\Vert }^2=⟨x,U^{\ast }Ux⟩=\Vert x{\Vert }^2$ and likewise $\Vert U^{\ast }x\Vert =\Vert x\Vert$, proving (4). Finally, if (4) holds, apply the same polarization identity to $U^{\ast }$ to obtain $U{U}^{\ast }=I$. This makes $U$ onto; the other norm identity supplies the remaining requirement of (1). ThusThese four statements are conditions in a single equivalence proof, rather than four independent source questions.

For $h,g\in Y$, their product $hg$ is holomorphic and integrable on the disk by the Cauchy-Schwarz inequality. Its angular mean on every centered circle is $h(0)g(0)$ by the mean value property. Polar integration therefore givesIf $g(0)=0$, this vanishes for every $h\in Y$, proving $\boxed{Z\subseteq Y^{\perp }}$. Integrability of the product justifies integration over the full disk, even when the individual holomorphic functions grow near its boundary.

The inclusion is strict. Set $w(z)=|z{|}^2-1/2$, a nonzero bounded function in $L^2(D)$. For every $h\in Y$,Thus $w\in Y^{\perp }$. It cannot equal $\overline{g}$ for holomorphic $g$: equality almost everywhere between these continuous functions would give equality everywhere, making $g=w$ a real-valued nonconstant holomorphic function, which is impossible. Therefore

The orthogonal complement isIt is a linear subspace and is closed, since it is an intersection of kernels of the continuous linear functionals $w\mapsto ⟨w,y⟩$. Continuity follows from the Cauchy-Schwarz inequality.

The minimizing point and orthogonality proved below write every $x$ as $x=y_0+(x-y_0)\in Y+Y^{\perp }$. If $w\in Y\cap Y^{\perp }$, then $\Vert w{\Vert }^2=⟨w,w⟩=0$, so the intersection is zero. ConsequentlyIn particular the minimizing point is unique: an alternative orthogonal decomposition would have a difference in this zero intersection.