Complete multiplicativity and $f(p)=\pm 1$ give, for $\mathrm{Re}s>1$,The product is nonzero there; its absolutely convergent logarithm has terms $f(p{)}^m{p}^{-ms}/m$. ThusEach numerator is one of $3\pm 4\cos \theta +\cos 2\theta$, so it is nonnegative. Exponentiation gives the required lower bound one.

If $F(1+it)=0$ with $t\ne 0$, its assumed holomorphy makes $|F(\sigma +it){|}^4=O((\sigma -1{)}^4)$, or smaller for a higher-order zero. The zeta factor at real $\sigma$ grows at most as $(\sigma -1{)}^{-3}$, while the zeta factor near $1+2it$ stays bounded. The product tends to zero, a contradiction. Hence

The double-angle formula givesFor $\sigma >1$, the logarithm of the absolutely convergent Euler product givesExponentiation provesSuppose $\zeta (1+it)=0$ for $t\ne 0$, of order $r\ge 1$. As $\sigma \downarrow 1$, its middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$ because the pole at one is simple, and the final factor is bounded because $1+2it\ne 1$. The product is $O((\sigma -1{)}^{4r-3})\to 0$, contradicting the lower bound one. ThereforeThis is the three-four-one zero-free-region argument specialized to the line with real part one.

The Möbius function satisfies $\mu (p^e{)}^2=1$ for $e=0,1$ and is zero for $e\ge 2$. ThereforeThe inverse local factor is $(1+p^{-s}{)}^{-1}={\sum }_{e\ge 0}(-1{)}^e{p}^{-es}$. Consequently the inverse under Dirichlet convolution is the Liouville functionwhere $\Omega (n)$ counts prime factors with multiplicity. Both functions are multiplicative functions. At $p^e$ with $e\ge 1$, their convolution is $(-1{)}^e+(-1{)}^{e-1}=0$, while its value at one is one. Multiplicativity then proves $h_1\ast h_2=I$ on every positive integer. Its Dirichlet series is $H_2(s)=\zeta (2s)/\zeta (s)$ for $\mathrm{Re}s>1$.

For a finite set of primes $p\le y$, multiply the absolutely convergent local series. Unique prime factorization and multiplicativity, with $g(1)=1$, giveThe omitted integers all exceed $y$, so the absolute difference from $G(s)$ is at most ${\sum }_{n>y}|g(n)|n^{-\sigma }$, which tends to zero. Hence the finite products converge to $G(s)$:This proves the Euler product directly from absolute convergence, rather than treating an infinite formal multiplication as automatically valid.

The Dirichlet convolution is $(f_1\ast f_2)(n)={\sum }_{d\mid n}{f}_1(d)f_2(n/d)$. Fix $s$ with $\sigma =\mathrm{Re}s>1$. Absolute convergence of the two Dirichlet series givesThus the double series may be regrouped by $n=ab$, provingThe same estimate proves absolute convergence of the resulting Dirichlet series.

The lower bound in (ii) makes the image of $I+T$ closed: if $(I+T)x_n$ converges, the bound makes $(x_n)$ Cauchy; completeness gives its limit and hence a image preimage. The same bound applies to $I+T^{\ast }$ by (iii), so its kernel is zero. The orthogonal complement identityshows that the image is dense, using the orthogonal decomposition by a closed subspace for its closure. Being both closed and dense, it equals $X$. Thus $I+T$ is onto and bounded below, and by part 3(a)(iii) it has a bounded inverse. Moreover

By the definition of the adjoint operator and conjugate symmetry,Taking real parts proves

A separate expansion givesHence $\boxed{\Vert (I+T)x\Vert \ge \Vert x\Vert }$. In particular $I+T$ is injective and bounded below by one.

Expanding both squared norms givesBoth norms are nonnegative, so taking square roots proves $\boxed{\Vert (I+T)x\Vert \ge \Vert (I-T)x\Vert }$.

Put $w=x-y_0$. For any $y\in Y$ and any real $t$, minimality givesThe derivative at zero must vanish, so $\mathrm{Re}⟨w,y⟩=0$. In a complex Hilbert space, apply the same argument to $iy$ to get the imaginary part zero as well. In the real case the first conclusion is already enough. Hence

Let $d={\inf }_{y\in Y}\Vert x-y\Vert$ and choose $y_n\in Y$ with $\Vert x-y_n\Vert \to d$. The parallelogram law givesThe right side tends to zero. Thus $(y_n)$ is Cauchy. The closed subspace $Y$ of a Hilbert space is complete, so $y_n\to y_0\in Y$. Continuity of the norm gives $\Vert x-y_0\Vert =d$, proving the distance is attained.

Evaluation at zero is a bounded linear functional for the uniform norm. DefineIt is linear and satisfies $|\varphi (f)|\le \sqrt{2\pi }\Vert T\Vert \Vert f{\Vert }_{\infty }$. Since $e_n(0)=(2\pi {)}^{-1/2}$,The factor $\sqrt{2\pi }$ is required by the source's basis normalization.

Since every $e_n$ is nonzero, boundedness givesCancel its norm to obtain $\boxed{\underset{n\in \mathbb{Z}}{\sup }|{\alpha }_n|\le \Vert T\Vert }$. Thus the multiplier sequence is bounded.

The normalization givesThus ${\alpha }_n{e}_n$ has exactly the coefficients prescribed for $T{e}_n$. Uniqueness yieldsThese are eigenvectors, including the zero-eigenvalue cases.

For each fixed $n$, the coefficient functional is bounded sinceSuppose $f_j\to f$ and $T{f}_j\to g$ uniformly. Taking coefficients and passing to the limit givesfor every fixed $n$. No uniform bound on ${\alpha }_n$ is needed for this passage. Uniqueness gives $g=Tf$, so $T$ has a closed graph. The space $X$ is Banach, being the closed subspace defined by $f(-\pi )=f(\pi )$ inside the continuous functions with the uniform norm. The closed graph theorem therefore proves $T$ is bounded.

For scalars $a,b$ and $f,h\in X$, linearity of the Fourier coefficients givesThe assumed uniqueness of the function with these coefficients implies $T(af+bh)=aTf+bTh$. Thus $T$ is linear.

The factorizationhas commuting factors because both are polynomials in $T$. If ${\lambda }^n\notin \sigma (T^n)$, their product is invertible. The preceding part would then make $T-\lambda I$ invertible, contradicting $\lambda \in \sigma (T)$. ThereforeThis proves the required inclusion without assuming the full spectral mapping theorem.

If $T$ has a bounded inverse, it is onto and $\Vert x\Vert \le \Vert T^{-1}\Vert \Vert Tx\Vert$, giving the claimed lower bound. Conversely a positive lower bound makes $T$ injective. Together with surjectivity it gives an everywhere-defined linear inverse with norm at most $1/\epsilon$. Hence it is bounded, proving the equivalence.

If $R,S$ are individually invertible, their product has inverse $S^{-1}{R}^{-1}$. Conversely assume $RS=SR$ and put $A=RS$, with $A$ invertible. Both factors commute with $A$ and therefore with $A^{-1}$. Direct multiplication givesThese are bounded two-sided inverses, so an invertible commuting product has both factors invertible. The commutation assumption is used when verifying the second side of each inverse.

The bounded operators on a Banach space form a Banach algebra in the operator norm. If $\Vert T\Vert <1$, the Neumann series $A={\sum }_{k=0}^{\infty }{T}^k$ converges absolutely in that norm. Its partial sums satisfyTaking limits proves $\boxed{(I-T{)}^{-1}=\sum _{k\ge 0}{T}^k}$.

If $|\lambda |>\Vert T\Vert$, factor $\lambda I-T=\lambda (I-T/\lambda )$ and apply the same argument. Thus such $\lambda$ are in the resolvent set and

For a bounded linear operator on a complex Banach space, its spectrum isThe complement is the resolvent set. For a real space, one may use its complexification when speaking of the complex spectrum; the algebraic arguments below apply over the underlying scalar field as well.