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A subset $B$ of a topological vector space is bounded when every neighborhood $U$ of zero absorbs it: $B\subseteq tU$ for all sufficiently large positive $t$. In a locally convex space defined by seminorms $p_{\lambda }$, this is equivalent to ${\sup }_{x\in B}{p}_{\lambda }(x)<\infty$ for every $\lambda$. In particular, a bounded subset of the Schwartz space has a uniform bound for every Schwartz seminorm; it need not have uniformly bounded supports.

For topological spaces $X,Y$, the compact-open topology on $C(X,Y)$ has subbasic setswhere $K\subseteq X$ is compact and $U\subseteq Y$ is open. Finite intersections of such conditions describe neighborhoods of a continuous map. This definition does not require a metric on either space.

For a metric target and continuous functions, it is the topology of uniform convergence on compact subsets. In the scalar case, the compact seminorms $p_K(f)={\sup }_{x\in K}|f(x)|$ describe it. If $X$ is a locally compact space and a Hausdorff space, this compact control also agrees with locally uniform convergence, because each point has a compact neighborhood.

On $\mathbb{R}$, control on every interval $[-n,n]$ is sufficient. It permits functions such as translated compactly supported bumps to converge to zero on every compact set while retaining the same global supremum norm. The compact-open topology records local approximation rather than a bound over an unbounded domain all at once.

The continuous dual space of a Hausdorff locally convex space separates points: for every nonzero $x$ there is a continuous linear functional $f$ with $f(x)\ne 0$.

A separating family of continuous seminorms supplies $p(x)>0$. In the real case, define $g(tx)=tp(x)$ on the span of $x$ and extend it under $p$ by the Hahn-Banach theorem. The bound $|g(y)|\le p(y)$ gives continuity. In a complex vector space, apply the real argument to the underlying real vector space, then use $f(y)=g(y)-ig(iy)$ to obtain a complex-linear functional with nonzero real part at $x$. The Hausdorff condition is essential: a nontrivial vector invisible to every continuous seminorm is also invisible to every continuous linear functional.

On ${\mathbb{R}}^2$, the coordinate seminorms $p_x$ and $p_y$ measure horizontal and vertical components separately. Their kernels of seminorms are the vertical and horizontal axes. Neither is a norm, but together they form a separating family of seminorms. A basic zero neighbourhood isEach condition alone is an unbounded strip; their intersection is an open rectangle. For $a=b=1$, this is the unit open ball of the maximum norm $\Vert (x,y){\Vert }_{\infty }=\max \{|x|,|y|\}$. For unequal tolerances it is the unit open ball of the weighted norm $\max \{|x|/a,|y|/b\}$. Thus the two coordinate measurements generate the usual topology of the plane.

For a numerical check, take $a=1$, $b=1/2$: $(0.8,0.4)$ passes both tests, $(0.8,0.7)$ passes only the horizontal test, and $(1.2,0.4)$ passes only the vertical test. A direction missed by one seminorm is detected by the other.

Here $C^{\infty }([0,1])$ means the real- or complex-valued functions continuous on $[0,1]$, smooth on $(0,1)$, with every interior derivative having a continuous extension to the endpoints; endpoint values of derivatives are these extensions. On this space of smooth functions on a compact interval, defineThese finite-valued seminorms increase with $k$ and separate points. In fact every $p_k$ is a norm, since it includes $p_0(f)=\Vert f{\Vert }_{\infty }$. The terminology seminorm includes norms; it does not assert that every measurement has a nontrivial kernel of a seminorm. Smooth convergence means $p_k(f_n-f)\to 0$ for every fixed $k$, equivalently uniform convergence of every derivative separately. It does not demand one tolerance or one convergence index that works for all orders simultaneously. The maximum convention is equivalent to the derivative supremum norm convention using a finite sum, since $p_k(f)\le {\sum }_{j=0}^k\Vert f^{(j)}{\Vert }_{\infty }\le (k+1)p_k(f)$.

For a worked example, $f(t)=t^3$ gives $p_0(f)=1$, $p_1(f)=3$, $p_2(f)=6$, and $p_k(f)=6$ for $k\ge 3$. The linear functional $\ell (f)=f^{\prime \prime }(1/2)$ obeys $|\ell (f)|\le p_2(f)$ and is continuous. Similarly, differentiation is a continuous linear operator $D:f\mapsto f^{\prime }$ for smooth convergence, because $p_k(Df)\le p_{k+1}(f)$.

The space of smooth functions on a compact interval is complete for this seminorm topology. A Cauchy sequence has uniform limits $g_j$ for its derivatives of each order. Passing to the limit in the fundamental theorem of calculus identitygives $g_j(t)-g_j(0)={\int }_0^t{g}_{j+1}(s)ds$, so $g_j^{\prime }=g_{j+1}$ and $g_0$ belongs to the same space. This is the completeness of the smooth-function metric argument; the metric there and the increasing-$p_k$ metric above induce the same smooth topology and the same Cauchy sequences.

Given a bilinear pairing $⟨\cdot ,\cdot ⟩$ and a seminorm $p$, its dual seminorm isIt may be infinite when $y$ does not annihilate the kernel of $p$. Whenever it is finite, the defining inequality gives $|⟨x,y⟩|\le p(x)p^{\ast }(y)$.

For a test-function seminorm on a finite set, define the dual functionalIf the tests span the whole function space, the original seminorm is a norm, and this is its finite dual norm. Without that spanning condition it can be infinite. Indeed, a component of $g$ orthogonal to the test span pairs nontrivially with a direction on which the original seminorm vanishes; arbitrary multiples of that direction remain admissible. The dual functional is finite precisely on the test span.

For a closed, convex, symmetric and uniformly bounded test family $\mathcal{F}$, the Bipolar theorem for a dual pair identifies the dual unit ball with $\mathcal{F}$, including the lower-dimensional case. Submultiplicativity under pointwise products then controls powers of a test in a polynomial, the step used by the dense model theorem for a multiplicative test family.

The dual polytope consists of the real linear functionals $\alpha$ on $H_2(Y,\partial Y;\mathbb{R})$ such that $|\alpha (u)|\le \Vert u{\Vert }_T$ for every $u$. Thus it is the polar of the unit ball of the Thurston norm and annihilates that seminorm's kernel.

For the angular-frequency Fourier transform and Schwartz seminorms $p_{m,n}(\varphi )={\sup }_x|x^m{\varphi }^{(n)}(x)|$, every output seminorm is bounded by finitely many input seminorms. To prove this, differentiation under the integral sign and integration by parts giveThe generalized Leibniz rule, $\Vert \hat{g}{\Vert }_{\infty }\le \Vert g{\Vert }_1$, and $\Vert x^a{\varphi }^{(b)}{\Vert }_1\le \pi (p_{a,b}+p_{a+2,b})$ implyThis proves continuity of the Fourier transform on the Schwartz space and rapid decay of every transformed derivative.

The Fourier transform with negative exponential and inverse factor $(2\pi {)}^{-n}$ is a continuous isomorphism of the Schwartz space. Weighted frequency derivatives are transforms of derivatives of weighted input functions, so their suprema are controlled by finitely many Schwartz seminorms. Fourier inversion gives ${\mathcal{F}}^2=(2\pi {)}^n\mathcal{R}$, where $\mathcal{R}f(x)=f(-x)$. Transposition gives a continuous isomorphism on tempered distributions as well.

For a positive integer $k$ and a function $f:G\to \mathbb{C}$ on a finite abelian group, with uniform expectations, definewhere $\mathcal{C}$ is complex conjugation. The alternating conjugations measure agreement around additive cubes. The expression is nonnegative; for $k\ge 2$ its $2^k$th root is a norm. At $k=1$ it is only the seminorm $|\mathbb{E}f|$.

With the multiplicative derivative ${\partial }_{h}f(x)=f(x+h)\overline{f(x)}$,Repeated multiplicative derivatives remove low-degree polynomial phases: a unit-modulus phase of degree less than $k$ has $U^k$ norm one. The Gowers U2 norm is the fourth moment of Fourier coefficients on a finite abelian group and detects linear correlations; the Gowers U3 norm controls four-term arithmetic progression counts through the generalized von Neumann inequality for four-term progressions.^[ref]

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

The kernel of a seminorm is its zero set, a linear subspace by absolute homogeneity and the triangle inequality. The estimate $|p(x)-p(y)|\le p(x-y)$ shows that adding an element of $\ker p$ does not change $p$. Hence $\Vert x+\ker p\Vert =p(x)$ defines a norm on the quotient vector space $E/\ker p$.

On $E$ itself, $d_p(x,y)=p(x-y)$ is a pseudometric; it is a metric exactly when $p$ is a norm. In particular, zero distance in one pseudometric need not mean equality of vectors. A separating family of seminorms removes the common zero-distance ambiguity.

A locally convex space is a topological vector space with a neighbourhood basis at zero consisting of convex sets. Equivalently, its topology is a seminorm topology generated by some family $\mathcal{P}$ of seminorms. We allow non-Hausdorff spaces here; a separating family of seminorms gives precisely the Hausdorff case. Every normed vector space is locally convex, but smooth convergence generally requires a whole family of measurements.

The geometric and measurement descriptions fit together: seminorm sublevel sets are convex, while the Minkowski functional of an open, convex zero neighbourhood invariant under all scalar multiplications with $|\lambda |\le 1$ is a continuous seminorm. Such neighbourhoods can be chosen as a neighbourhood basis. ^[ref].

Local convexity makes Hahn-Banach theorem and separation of a point and an open convex set available. It turns geometric distinctions into continuous linear functionals, connecting convex sets to the continuous dual space. For a seminorm topology generated by $\mathcal{P}$, a linear functional $\ell$ is continuous exactly when there are $p_1,\dots ,p_m\in \mathcal{P}$ and $C>0$ withThus one continuous linear functional needs only finitely many of the available measurements. For example, point evaluation of a derivative on the space of smooth functions on a compact interval is controlled by a suitable derivative seminorm.

Let $X$ be a real Hausdorff topological vector space, and let $U$ be a bounded, convex, open neighbourhood of zero with $U=-U$. Its Minkowski functionalis a norm and $U=\{x:p_U(x)<1\}$. Convexity and symmetry give a real balanced set, so the gauge is a seminorm. Openness gives the displayed unit-ball identity.

To prove definiteness, suppose $p_U(x)=0$. For any zero neighbourhood $V$, boundedness of $U$ gives $\epsilon U\subseteq V$ for a sufficiently small $\epsilon >0$. Since $x\in \epsilon U$, it belongs to every zero neighbourhood. The Hausdorff property forces $x=0$. The same boundedness comparison shows that the norm topology agrees with the original topology.

For a complex vector space, require $U$ to be balanced under complex scalars; symmetry under minus one alone only supplies the real homogeneity. The argument is independent of finite dimension and requires no Euclidean inner or outer balls.^[ref]

A continuous surjective linear map between Fréchet spaces is an open map. The Baire category theorem first shows that the closure of the image of every convex balanced zero-neighbourhood is a zero-neighbourhood: the target is the countable union of scalar multiples of this image, and an interior point of its closure can be translated to zero using convexity and symmetry.

To remove the closure, fix a domain zero-neighbourhood $U$ and choose convex balanced zero-neighbourhoods $U_n$ so that every series ${\sum }_n{u}_n$, $u_n\in U_n$, converges to a point of $U$. With an increasing defining sequence of seminorms $p_j$, one can ensure $p_j(u_n)<2^{-n}$ for $j\le n$ and make the finitely many seminorms defining $U$ summable within its bounds. Completeness gives convergence. Choose target zero-neighbourhoods $V_n\subset \overline{A(U_n)}$ shrinking to zero. For $y\in V_1$, choose $u_n\in U_n$ successively so that $y-A{\sum }_{j\le n}{u}_j\in V_{n+1}$. Density permits each correction. Continuity then gives $y=A{\sum }_n{u}_n\in A(U)$, proving openness.

A distribution is a linear functional on the space of test functions that is continuous in its test-function topology. Equivalently, for every compact $K\subset \mathbb{R}$ there are an integer $m$ and a constant $C_K$ such that $|⟨T,\varphi ⟩|\le C_K{\max }_{0\le j\le m}\Vert {\varphi }^{(j)}{\Vert }_{\infty }$ whenever $\varphi$ has support in $K$.

Linearity of $T_n$ follows from linearity of the integral and derivative. Since the integration interval has length $2/n$,This is a test-function seminorm estimate of order one, valid for every $K$ and even uniformly in $n$. It proves that each $T_n$ is a distribution.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$, with complex-linear distribution pairings throughout. The Riemann-Lebesgue lemma says that, for an integrable function $f\in L^1(\mathbb{R})$, its Fourier transform is continuous and tends to zero as $|\xi |\to \infty$. It also satisfies $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. The product rule, or convolution theorem, isHere $(f\ast g)(x)=\int f(y)g(x-y)dy$ belongs to $L^1(\mathbb{R})$ by Tonelli's theorem, and Fubini's theorem proves the displayed identity. The companion Fourier product formula, for Schwartz functions, is $\widehat{fg}=(2\pi {)}^{-1}\hat{f}\ast \hat{g}$.

A Schwartz function is a smooth function $\varphi :\mathbb{R}\to \mathbb{C}$ for which every Schwartz seminormis finite. These seminorms define the topology of the Schwartz space $\mathcal{S}(\mathbb{R})$. In particular,Thus every Schwartz function is an integrable function.

The Fourier bounds for Schwartz functions assert that each Schwartz seminorm of $\hat{\varphi }$ is bounded by finitely many Schwartz seminorms of $\varphi$. Explicitly, differentiation under the integral sign and integration by parts giveso the generalized Leibniz rule and the preceding integrability estimate giveIn particular the Fourier transform maps the Schwartz space continuously into itself.

For a nonzero complex unital Banach algebra, defineThis is the spectrum of an element; its complement carries the resolvent of an element $R_x(\lambda )=(\lambda 1-x{)}^{-1}$.

If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges in $A$. Multiplication of finite partial sums followed by a norm limit gives both inverse identities. Thus the spectrum is bounded by $\Vert x\Vert$. The invertible set is open by the allowed assumption, and $\lambda \mapsto \lambda 1-x$ is continuous, so the spectrum is closed. It is therefore compact if nonempty.

To prove nonemptiness, suppose the resolvent exists for every $\lambda$. At a fixed ${\lambda }_0$, the same series applied locally givesfor $|h|\Vert R_x({\lambda }_0)\Vert <1$, proving norm analyticity. Every bounded complex linear functional $\psi \in A^{\ast }$ therefore gives an entire scalar function $\psi (R_x(\lambda ))$. On compact disks it is bounded by continuity of inversion, and the large-$\lambda$ series givesHere the factor $\Vert 1\Vert \ge 1$ handles an algebra norm whose identity was not normalized to one. The Liouville theorem forces each scalar function to be identically zero. The Hahn-Banach theorem separates points of $A$, so $R_x(\lambda )=0$, contradicting $(\lambda 1-x)R_x(\lambda )=1$. HenceThis includes an explicit proof of the nonemptiness of the Banach-algebra spectrum rather than using it as a named result.

For the algebra of entire functions, choose the compact-disk algebra norm on entire functionsIt is finite by continuity on a compact disk, satisfies the norm axioms, and is submultiplicative. If the norm is zero, $f$ vanishes on the open disk, so the identity theorem gives $f=0$ on the plane. Thus it is an actual algebra norm, not merely a seminorm.

No complete algebra norm is possible. For the coordinate function $Z(z)=z$, the function $Z-\lambda 1$ has a zero at $\lambda$ for every complex $\lambda$ and hence cannot have a multiplicative inverse among entire functions. Thus its algebraic spectrum is all of $\mathbb{C}$, independently of any chosen norm. Were an algebra norm complete, the spectral theorem just proved would make that spectrum bounded, a contradiction. Therefore entire functions admit an algebra norm but no complete algebra norm. This distinguishes failure of every complete algebra norm from failure of just the displayed disk norm.

For continuous functions on the plane, construct disjoint plateaux explicitly. LetThe supports of ${\chi }_n$ lie in the disks $|z-n|\le 1/4$, so they are pairwise disjoint and form a locally finite family. Thus $f$ is continuous. On the support where $g_n$ is nonzero, ${\chi }_n=1$ and all other bumps vanish, giving $f{g}_n=n{g}_n$; also $g_n(n)=1$, so no $g_n$ is zero. If an algebra norm existed, submultiplicativity would implyso $n\le \Vert f\Vert$ for every positive integer, impossible. Consequently the algebra of all continuous functions on the plane admits no algebra norm, with no completeness hypothesis needed.

Write $K={\sigma }_A(x)$ and $R(\lambda )=(\lambda 1-x{)}^{-1}$. Choose a bounded finite union of small polygonal regions $V$ with $K\subset V$, $\overline{V}\subset U$ and $\partial V\cap K=\varnothing$. Its positively oriented boundary cycle $\Gamma$ has winding number one on $K$ and zero outside $V$; boundaries of holes have negative orientation. Such regions can be obtained from a sufficiently fine grid around the compact set $K$. This accommodates disconnected spectra and neighborhoods with holes.

Define the holomorphic functional calculus by the norm-convergent contour integralThe integrand is continuous on a finite rectifiable cycle, so the integral exists in the Banach space $A$. Its value is independent of the chosen cycle: on $U\setminus K$ the integrand is holomorphic, and cycles with the stated winding numbers are homologous there. Apply the scalar Cauchy integral theorem to every bounded linear functional and then use the Hahn-Banach theorem to recover equality of the algebra-valued integrals. Equivalently, a common refinement of the polygonal regions gives the same contour-deformation argument.

Linearity is immediate. For the entire functions $1,Z$, enlarge the integration cycle to a sufficiently large circle in the resolvent set, using their entire extension. Integrating the uniformly convergent Neumann series term by term givesTo prove multiplicativity, choose two nested admissible cycles, ${\Gamma }_o$ outside ${\Gamma }_i$, both surrounding $K$ and lying in $U$. The resolvent identity givesUse ${\Gamma }_o$ for $f$ and ${\Gamma }_i$ for $g$, multiply the two integrals, and insert this identity. In the term with $R(\mu )$, the outer Cauchy integral formula replaces $f(\lambda )/(\lambda -\mu )$ by $f(\mu )$. In the term with $R(\lambda )$, the inner integral of $g(\mu )/(\lambda -\mu )$ is zero because $\lambda$ is outside the inner region. ThusThe exchanges of integrals are justified by continuous integrands on compact separated cycles. This proves the unital algebra-homomorphism property, not just an identity of algebra character values.

For continuity in local uniform convergence, fix one cycle and boundThe last supremum is a defining compact-set seminorm of $\mathcal{O}(U)$. Hence the map is continuous.

If another continuous unital algebra homomorphism $\Psi$ sends $Z$ to $x$, its values on polynomials are forced. For every $\zeta \notin U$, the function $(Z-\zeta {)}^{-1}$ exists in $\mathcal{O}(U)$, so multiplicativity forcesTherefore $\Psi$ and ${\Theta }_x$ agree on all rational functions with poles outside $U$. The permitted Runge theorem makes these functions dense in $\mathcal{O}(U)$ for local uniform convergence, including when $U$ is disconnected. Continuity makes the maps agree everywhere. ThusThis proves the continuity and uniqueness of holomorphic functional calculus with the exact topology required.

For an algebra character $\varphi$, multiplicativity and nonzeroness give $\varphi (1)=1$. An invertible element cannot have zero algebra character value, so $\varphi (x)\in {\sigma }_A(x)$. The same observation for each algebra element and the spectral bound from Question 3 gives $|\varphi (b)|\le \Vert b\Vert$; thus a algebra character is automatically a bounded linear functional. In particular,Apply $\varphi$ inside the contour integral and use the Cauchy integral formula with winding number one at $\varphi (x)$:No separation of algebra elements by algebra characters is used; such separation can fail for a general commutative Banach algebra.

For the final clause, the two closed spectral pieces $X,Y$ are compact and have positive distance. Choose disjoint open neighborhoods $U_X,U_Y$ and let $f$ be identically one on $U_X$ and zero on $U_Y$. This locally constant function is holomorphic on their union, and $f^2=f$. Set $e={\Theta }_a(f)$. Multiplicativity gives $e^2=e$, and the proved algebra character formula givesThis is the spectral idempotent from a separated spectrum. It is also nontrivial without assuming that algebra characters separate points. If $e=0$, choose $\lambda \in X$ and define $g=0$ on $U_X$, $g(z)=1/(z-\lambda )$ on $U_Y$. Then $(Z-\lambda )g=1-f$, so $(a-\lambda 1){\Theta }_a(g)=1$, with commuting factors, making $a-\lambda 1$ invertible despite $\lambda \in {\sigma }_A(a)$. This is impossible. If $e=1$, exchange the two pieces and use $\lambda \in Y$ for the analogous contradiction. Hence $e\ne 0,1$.