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For a real-valued loss function depending only on evaluations at $x_1,\dots ,x_n$, and a positive squared-norm penalty, any minimizing tuple in a sum of reproducing-kernel Hilbert spaces has the formwith the same coefficient vector for every component. The tuple must be the unique minimum-norm decomposition of its sum. Apply the representer theorem to that sum and note that the displayed tuple has total squared norm ${\sum }_j{\hat{\alpha }}^T{K}_j\hat{\alpha }={\hat{\alpha }}^T({\sum }_j{K}_j)\hat{\alpha }$, exactly the norm of the sum. This argument is conditional on existence of a minimizer; arbitrary loss functions need not attain their infimum.

A Cauchy sequence $(u_n)$ in the BV space converges in $L^1$ to $u$. For fixed sufficiently large $n$, lower semicontinuity gives $|D(u_n-u)|(\Omega )\le {\mathrm{liminf}}_m|D(u_n-u_m)|(\Omega )$. The $L^1$ norm of the same difference converges, so the small full-norm Cauchy bound passes to $u_n-u$. The triangle inequality then puts $u$ in the BV space and proves convergence in its full norm. This argument turns completeness of $L^1$ and lower semicontinuity of variation into completeness of the combined norm.

The completion of a normed vector space $X$ is a Banach space containing an isometric dense copy of $X$. Construct it from Cauchy sequences modulo sequences whose difference tends to zero; define the norm by $\Vert [(x_n)]\Vert ={\lim }_n\Vert x_n\Vert$. Equivalently, the canonical evaluation map into the bidual space is an isometry by the Hahn-Banach theorem, and the closure of its image supplies a completion.

A Banach space is precisely a normed vector space in which every Cauchy sequence converges in norm to an element of the same space. Completion of a normed space fills missing limits without changing distances between existing elements. For example, $C([0,1])$ with the L2 norm completes to L2 space on the interval with Lebesgue measure. The bounded integral functional extends uniquely to that completion of a normed space: define $F(f)={\lim }_{k}F(f_k)$ for any approximating sequence of continuous functions. Its bound makes this independent of the approximation. The unbounded point evaluation has no continuous extension agreeing with its values on all continuous functions.^[ref]

The squared distance to a convex set is the infimal convolution of the set's indicator functional of a constraint set and the half-squared Hilbert space norm. Its convex conjugate is the sum of the half-squared norm and the support function of the set.

For $T>0$ and the constant matrices over the real numbers in the Kalman rank criterion, the finite-time Controllability Gramian accumulates the influence of all command times. Its column space equals that of the controllability matrix, and $W_T$ is a symmetric positive-definite matrix exactly when the controllability matrix has rank of a matrix $n$. If so, put $\delta =x_1-e^{AT}{x}_0$. Among square-integrable functions steering $x_0$ to $x_1$ at $T$, the commandminimizes ${\int }_0^T\Vert u(t){\Vert }^{2}dt$, with value ${\delta }^T{W}_T^{-1}\delta$. Here $T$ in a matrix superscript denotes transpose. This cost is a squared input norm, not necessarily physical energy, and the optimizing command need not satisfy an amplitude bound. ^[ref]

For the double integrator, $e^{As}B=(s,1{)}^T$, so direct integration givesFor a rest-to-rest distance $d$, $u_{\ast }(t)=6d/T^2-12dt/T^3$ and the minimum squared input norm is $12d^2/T^3$. Compare the two constant commands in the double integrator example, whose cost is $16d^2/T^3$: both attain the target, but the smooth command costs less. At $T=2$, $d=1$, however, $u_{\ast }$ peaks at $3/2$, so it violates $|u|\le 1$ while the two constant commands remain admissible. Controllability, optimal control, and constrained reachability in control theory answer different questions.

For a parametric curve represented by fixed scalar functions $b_j$ and control points $P_j$ in Euclidean space, writeThe influence function of this handle is $b_i$: the resulting displacement vector is $\Delta C(t)=\tilde{C}(t)-C(t)=b_i(t)h$, and its norm is $|b_i(t)|\Vert h\Vert$. These are exact finite changes, not only a first-order estimate. Where $b_i$ is negative the displacement points opposite to $h$. Where $b_i(t)=0$, the curve is unchanged at that parameter.

For the cubic Bézier curve above, the Bernstein basis influence function is $b_{1,3}(t)=3t(1-t{)}^2$. Its derivative isIt increases up to $t=1/3$ and decreases thereafter on $[0,1]$. Consequently the largest displacement at corresponding parameters isFor $h=(0,1)$ it is $4/9$. This is a comparison at the same parameter. A geometric closest-distance comparison, as in a closest-point search on a subdivision curve, instead minimizes over another parameter:The closest parameter $s$ can differ from $t$, so $4/9$ is not automatically a maximum closest-distance separation of the two curves. Neither parameter is assumed to be arc length.

The support of a function $b_i$ is the closure of the parameter values where $b_i\ne 0$. It bounds where a nonzero handle move can act, but may include boundary points where the actual displacement is zero. Here the support of a function $b_{1,3}$ is all of $[0,1]$ relative to the parameter domain, while the endpoints are unaffected. In contrast, a B-spline of order $k$ has support within $[t_i,t_{i+k}]$ in its spline knot sequence, so its handle affects only that parameter range; the support and Gram bandwidth of B-splines describes the same locality in another application. Changing the spline knots or the basis along with the handle requires a different comparison.

When the basis functions form a partition of unity, ${\sum }_j{b}_j(t)=1$, each evaluation is an affine combination of the control points. Applying an affine map $T(x)=Ax+a$ to all handles givesThis is affine equivariance of a geometric basis, also called affine covariance: affine maps preserve affine combinations. In particular, translating every handle by $h$ translates the whole curve by $h$, whereas moving just one handle produces $b_i(t)h$. The Bernstein basis has this partition of unity property.^[ref] Nonnegative weights further make evaluations convex combinations, giving the convex hull bound illustrated by the control polygon; nonnegativity is not needed for affine covariance.

For a proof outline of the influence formula, subtract the two basis representations: every unchanged coefficient cancels, leaving $b_i(t)(P_i+h)-b_i(t)P_i=b_i(t)h$. The same subtraction with several moved handles gives $\Delta C(t)={\sum }_i{b}_i(t)h_i$. No claim about interpolation, closest-point search on a subdivision curve or endpoint motion follows until the actual basis values are examined.

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.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

The continuous functional calculus of a bounded normal operator $T$ is a representation of $C(K)$, with $K=\sigma (T)$. The Riesz-Markov-Kakutani representation theorem represents $f\mapsto ⟨f(T)v,v⟩$ by a positive regular measure ${\mu }_v$. The assignment $f\mapsto f(T)v$ is isometric for the $L^2({\mu }_v)$ norm, and extends to a unitary map onto the closed cyclic subspace it generates. Multiplication by $z$ and $\overline{z}$ represent $T$ and $T^{\ast }$. Orthogonal decomposition into cyclic reducing subspaces, allowing an uncountable family, constructs the Borel functional calculus for a normal operator without requiring separability.

Definiteness says that a norm vanishes only at the zero vector. The box norm on a nonempty finite Cartesian product is definite because its defining sum of squares includes equal-column terms, each of which is the square of an average of squares. Omitting repeated coordinates would destroy this proof and can produce a different quantity.

A continuous bilinear map obeys $\Vert f(h,k)\Vert \le C\Vert h\Vert \Vert k\Vert$. Expanding $f(a+h,b+k)$ leaves the displayed linear term and remainder $f(h,k)$. For product norm $\rho =(\Vert h{\Vert }^2+\Vert k{\Vert }^2{)}^{1/2}$, the remainder has norm at most $C{\rho }^2/2$, hence is $o(\rho )$. This proves the Fréchet derivative formula.

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.