There is Continuous Function That Vanishes Bounded Set

Interpolation of Operators

In Pure and Applied Mathematics, 1988

Definition 4.2

The space of bounded continuous functions on R ⁿ is denoted by C = C(R ⁿ ). It is a Banach space under the supremum norm f → ||f||_∞ = sup{|f(x)|: x ∈ R ⁿ }.

The r-th order modulus of continuity of a function f in L^p (R ⁿ ), 1 ≤ p < ∞, is defined by

(4.2) $\begin{array}{cc}{\omega }_r{(f,t)}_p=\underset{\left|h\right|\le t}{\sup }\Vert {\Delta }_h^{r}f\Vert p,& (f\in L^p,t>0).\end{array}$

When p = ∞, the L^p -norm is replaced by the norm in C:

(4.3) $\begin{array}{ll}{\omega }_{\text{r}}{(f,t)}_{\infty }=\underset{\left|h\right|\le t}{\sup }{\Vert {\Delta }_h^{r}f\Vert }_{\infty },\hfill & (f\in C,t>0)\hfill \end{array}.$

Each modulus ω _r(f, t)_p, (1 ≤ p ≤ ∞), is a nonnegative increasing function of t > 0; furthermore, for each fixed t, ω _r (·-,t)_p is a seminorm on L^p or C. It follows from (4.1) that

(4.4) ${\omega }_{\text{r}}{(f,t)}_p\le 2^r{\Vert f\Vert }_p.$

Since Δ_2h = T ² _h — I = (T_h + I)Δ _h, one sees also that

(4.5) ${\omega }_r{(f,2t)}_p\le 2^r{\omega }_r{(f,t)}_p.$

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Eigenvalues in Riemannian Geometry

In Pure and Applied Mathematics, 1984

Theorem 5

If f is a bounded continuous function on R ⁿ , then u(x, t) given by (16) is the unique solution to the initial-value problem for the heat equation (1): (7) among all functions v(x, t) on R ⁿ × (0, ∞) for which $\left|\upsilon \left(x,t\right)\right|\le \text{const},\left|\left({\text{grad}}_{x}u\right)\left(x,t\right)\right|\le \text{const}/\sqrt{t},$ where the constants are independent of (x, t).

Finally, we consider a restricted case of the nonhomogeneous initial-value problem.

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Time-invariant Linear Systems

R.F. Hoskins Research Professor , in Delta Functions (Second Edition), 2011

4.2.2 Convolution integral

Now take for the input x a bounded, continuous function which vanishes identically outside some finite time-interval, say α  ≤ t  ≤ β. We can approximate x(t) by a train of adjacent narrow pulses as illustrated in Fig. 4.4,

Figure 4.4.

$x\left(t\right)\simeq {\displaystyle \sum _{k}x\left({\tau }_k\right)\left[u\left(t-t_k\right)-u\left(t-t_{k-1}\right)\right]}\text{.}$

The response of the system to an elementary pulse of width Δτ_k , where Δτ_k  =   t_k   − t _{k  − 1} , and of height x(τ_k ), centred about the point t   =   τ_k , will be given (approximately) by

$\left\{x\left({\tau }_k\right)\Delta {\tau }_k\right\}h\left(t-{\tau }_k\right)\text{,}$

using (4.8), and the fact that the system is time-invariant. Next, by linearity of the system, the response to the input x(t) will be given, again to a first approximation, by the sum

(4.9) ${\displaystyle \sum _{k}x\left({\tau }_k\right)h\left(t-{\tau }_k\right)\Delta {\tau }_k\text{.}}$

In the limit, as the approximating pulses are made narrower and narrower, we obtain the actual output у in the form of a so-called convolution integral

(4.10) $y\left(t\right)={\displaystyle {\int }_{\alpha }^{\beta }x\left(\tau \right)h\left(t-\tau \right)\mathit{d\tau }={\displaystyle {\int }_{-\infty }^{+\infty }x\left(\tau \right)h\left(t-\tau \right)\mathit{d\tau }\text{.}}}$

We can take infinite limits in (4.10) because the function x is known to vanish identically outside the finite interval [α, β]. If we remove this constraint on x then an extension of the above argument can be used to show that the output у will still be given by an integral of the form (4.10), provided that the functions x and h are suitably well-behaved for large values of |t| (so that the infinite integral does converge). Similarly we may weaken considerably the assumption that x and h are continuous. Note, however, that the one crucial step in the argument is the passage to the limit in the derivation of (4.10) from (4.9). It is this step which relies explicitly on a suitable assumption of system continuity as discussed in the Remarks at the end of section 4.1 above.

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Modern General Topology

In North-Holland Mathematical Library, 1985

Example III.12

All bounded subsets of a Euclidean space E ⁿ are totally bounded, and they are the only totally bounded subspaces of E ^n. On the other hand the closed sets and only those are complete subspaces of E ^n. Generally, we can prove that every generalized Hilbert space H(A) is a complete metric space (but non-compact). To show it we suppose

$\begin{array}{cc}{p}^{\left(i\right)}=\left\{p_{\alpha }^{\left(i\right)}|\alpha \in A\right\},& i=1,2,\dots ,\end{array}$

form a Cauchy point sequence of H(A). Since for each $\alpha ,\left\{p_{\alpha }^{\left(i\right)}|i=1,2,\dots \right\}$ is a Cauchy point sequence of E ¹, it converges to a point p _α of E ¹, i.e.,

Let ε > 0 be given; then there is i ₀ such that for every $i,i\text{'}⩾i_0,$

${\left({\displaystyle \sum _{\alpha \in A}{\left(p_{\alpha }^{\left(i\right)}-p_{\alpha }^{\left(i\text{'}\right)}\right)}^2}\right)}^{1/2}<\epsilon .$

Hence for every finite subset A' of A, we obtain

${\left({\displaystyle \sum _{\alpha \in A\text{'}}{\left(p_{\alpha }^{\left(i\right)}-p_{\alpha }^{\left(i\text{'}\right)}\right)}^2}\right)}^{1/2}<\epsilon .$

Letting i' → ∞ in this inequality, we obtain

(1) ${\left({\displaystyle \sum _{\alpha \in A\text{'}}{\left(p_{\alpha }^{\left(i\right)}-p_{\alpha }\right)}^2}\right)}^{1/2}⩽\epsilon .$

This implies that

${\left({\displaystyle \sum _{\alpha \in A\text{'}}{p}_{\alpha }^2}\right)}^{1/2}-{\left({\displaystyle \sum _{\alpha \in A\text{'}}{\left(p_{\alpha }^{\left(i\right)}\right)}^2}\right)}^{1/2}⩽{\left({\displaystyle \sum _{\alpha \in A\text{'}}{\left(p_{\alpha }^{\left(i\right)}-p_{\alpha }\right)}^2}\right)}^{1/2}⩽\epsilon ,$

i.e.,

$\left({\displaystyle \sum _{\alpha \in A\text{'}}{p}_{\alpha }^2}\right)⩽\epsilon +{\left({\displaystyle \sum _{\alpha \in A\text{'}}{\left(p_{\alpha }^{\left(i\right)}\right)}^2}\right)}^{1/2}.$

Since ${\displaystyle {\sum }_{\alpha \in A}{\left(p_{\alpha }^{\left(i\right)}\right)}^2}<+\infty ,\text{ we get}{\displaystyle {\sum }_{\alpha \in A}{p}_{\alpha }^2}<+\infty ,\text{ i}\text{.e}\text{.,}$

On the other hand, (1) implies

$\rho \left(p,p^{\left(i\right)}\right)={\left({\displaystyle \sum _{\alpha \in A}{\left(p_{\alpha }^{\left(i\right)}-p_{\alpha }\right)}^2}\right)}^{1/2}⩽\epsilon .$

Therefore $\left\{p^{\left(i\right)}|i=1,2,\dots \right\}$ converges to p, proving that H(A) is complete.

Let C ^*(X) be the collection of all bounded continuous functions over a topological space X. We introduce into C ^*(X) the metric given in Example III.9. Then we can easily see that C ^*(X) is a complete metric space. If X is the unit segment I = [0, 1], then it follows from Weierstrass' theorem ¹ that C ^*(I) (=C(I)) is also separable and therefore it satisfies the 2nd axiom of countability. However, it is not compact.

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Functionals and Representations

In C*-Algebras and their Automorphism Groups (Second Edition), 2018

3.10.2

If Q is a convex set in a real topological vector space, then we denote by $A(Q)$ the Banach space of all continuous bounded functions a on Q that are affine in the sense that

$a(\lambda \varphi +(1-\lambda )\psi )=\lambda a(\varphi )+(1-\lambda )a(\psi )$

for every convex combination of points ϕ and ψ in Q. We denote by $B(Q)$ the Banach space of all bounded affine functions on Q. If $0\in Q$ , then $A_0(Q)$ (respectively, $B_0(Q)$ ) denotes the set of elements in $A(Q)$ (respectively, $B(Q)$ ) that vanish at zero.

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Spectral Theory for Automorphism Groups

In C*-Algebras and their Automorphism Groups (Second Edition), 2018

8.14.2 Proposition

Let K be a subspace satisfying the conditions in 8.13.2 , and let Δ be the modular operator defined in 8.13.4 . Then the representation $t\to {\mathrm{\Delta }}^{\mathrm{i}t}$ is the unique unitary representation of $\mathsf{R}$ that satisfies the modular condition with respect to K.

Proof

Note that ${\mathrm{\Delta }}^{\mathrm{i}t}K=K$ by 8.13.5. Take $\xi ,\eta$ in K. Since $K\subset \mathcal{D}({\mathrm{\Delta }}^s)$ for $0⩽s⩽\frac{1}{2}$ , by 8.13.4 we can define a bounded continuous function f on ${\mathbf{\Omega }}_{-1}$ by

$f(t-\mathrm{i}s)=({\mathrm{\Delta }}^{\mathrm{i}t}{\mathrm{\Delta }}^{s/2}\xi |{\mathrm{\Delta }}^{s/2}\eta ),t-\mathrm{i}s\in {\mathbf{\Omega }}_{-1}.$

The function $\zeta \to {\mathrm{\Delta }}^{\mathrm{i}\zeta }\xi$ is holomorphic for $-\frac{1}{2}<\mathrm{Im}\zeta <0$ by the spectral theorem (with derivative $\mathrm{i}{\mathrm{\Delta }}^{\mathrm{i}\zeta }\log \mathrm{\Delta }\xi$ ). Since we can write any ${\zeta }_0$ in the interior of ${\mathbf{\Omega }}_{-1}$ in the form ${\zeta }_0=\zeta -\mathrm{i}r$ , where $-\frac{1}{2}<\mathrm{Im}\zeta <0$ and $0<r<\frac{1}{2}$ , whence $f({\zeta }_0)=({\mathrm{\Delta }}^{\mathrm{i}\zeta }\xi |{\mathrm{\Delta }}^r\eta )$ ; it follows that f is holomorphic in the interior of ${\mathbf{\Omega }}_{-1}$ . For ${\zeta }_0=t-\mathrm{i}$ , by 8.13.4, 8.13.5, and 8.13.3(iii) we obtain

$f(t-\mathrm{i})=({\mathrm{\Delta }}^{\mathrm{i}t}{\mathrm{\Delta }}^{1/2}\xi |{\mathrm{\Delta }}^{1/2}\eta )=({\mathrm{\Delta }}^{\mathrm{i}t}j\xi |j\eta )=(j{\mathrm{\Delta }}^{\mathrm{i}t}\xi |j\eta )=(\eta |{\mathrm{\Delta }}^{\mathrm{i}t}\xi ).$

Suppose now that $(u,H)$ was another unitary representation of $\mathsf{R}$ satisfying the modular condition with respect to K. If ${\xi }_1$ is an analytic vector for u in K, then the analytic function $\zeta \to u_{\zeta }{\xi }_1$ must satisfy

$(u_{t-\mathrm{i}}{\xi }_1|\eta )=(\eta |u_t{\xi }_1)$

for every η in K. Taking ${\eta }_1$ to be analytic for the group $t\to {\mathrm{\Delta }}^{\mathrm{i}t}$ , we must similarly have

$(\xi |{\mathrm{\Delta }}^{\mathrm{i}t+1}{\eta }_1)=({\mathrm{\Delta }}^{\mathrm{i}t}{\eta }_1|\xi )$

for every ξ in K.

Define the analytic function g by

$g(\zeta )=(u_{\zeta }{\xi }_1|{\mathrm{\Delta }}^{\mathrm{i}(\bar{\zeta }-\mathrm{i})}{\eta }_1).$

Since ${\mathrm{\Delta }}^{\mathrm{i}t}K=K$ , we have

$g(t-i)=(u_{t-\mathrm{i}}{\xi }_1|{\mathrm{\Delta }}^{\mathrm{i}t}{\eta }_1)=({\mathrm{\Delta }}^{\mathrm{i}t}{\eta }_1|u_t{\xi }_1).$

Similarly, since $u_{t}K=K$ , we get

$g(t)=(u_t{\xi }_1|{\mathrm{\Delta }}^{\mathrm{i}t+1}{\eta }_1)=({\mathrm{\Delta }}^{\mathrm{i}t}{\eta }_1|u_t{\xi }_1).$

It follows that $g(\zeta -\mathrm{i})=g(\zeta )$ for all ζ, since it is true for all ζ in $\mathsf{R}$ . The functions $\zeta \to u_{\zeta }{\xi }_1$ and $\zeta \to {\mathrm{\Delta }}^{\mathrm{i}\zeta }{\eta }_1$ are bounded on horizontal strips; in particular, g is bounded on ${\mathbf{\Omega }}_{-1}$ . The periodicity of g implies that g is bounded on $\mathsf{C}$ and therefore constant by Liouville's theorem. Consequently,

$({\mathrm{\Delta }}^{\mathrm{i}t}{\eta }_1|u_t{\xi }_1)=g(t)=g(0)=({\eta }_1|{\xi }_1).$

Since this is valid for a dense set of vectors in K and $K+\mathrm{i}K$ is dense in H, we conclude that $u_{-t}{\mathrm{\Delta }}^{\mathrm{i}t}=1$ for all t, as desired. □

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The Dirac Delta Function

R.F. Hoskins Research Professor , in Delta Functions (Second Edition), 2011

2.4.1 Definition of Riemann-Stieltjes integral

An entirely different way of representing the sampling operation associated with the so-called delta function is to resort to a simple generalisation of the concept of integration itself. Recall that the elementary (or Riemann) theory of the integration of bounded, continuous functions over finite intervals treats the integral as the limit of finite sums

${\displaystyle {\int }_a^{b}f\left(t\right)\mathit{dt}=\underset{\mathit{\Delta t}\to 0}{lim}{\displaystyle \sum _{k=1}^{n}f\left({\tau }_k\right)\left(t_k-t_{k-1}\right)}\text{.}}$

Here the t_k are points of subdivision of the range [a, b] of integration, the τ_k are arbitrarily chosen points in the corresponding sub-intervals [t _k–1,t_k ], and Δt represents the largest of the quantities Δ _kt ≡ t_k – t _k–1. In a generalisation due to Stieltjes these elements Δ _kt are replaced by quantities of the form Δ _k ν ≡ ν(t_k ) –ν(t _k–1), where ν is some fixed, monotone increasing function, and the so-called Riemann-Stieltjes integral appears as the limit of appropriately weighted finite sums.

Let ν be a monotone increasing function and let f be any function which is continuous on the finite, closed interval [a, b]. By a partition P of [a, b] we mean a subdivision of that interval by points t_k, where 0 ≤ k ≤ n, such that

$a=t_0<t_1<\dots <t_{n-1}<t_n=b\text{.}$

For a given partition P let Δ _k ν denote the quantity

${\Delta }_k\nu \equiv \nu \left(t_k\right)-\nu \left(t_{k-1}\right),k=1,2,\dots ,n\text{.}$

Then, if τ_k is some arbitrarily chosen point in the sub-interval [t _k–1, t_k ], where 1 ≤ k ≤ n, we can form the sum

(2.10) ${\displaystyle \sum _{k=1}^{n}f\left({\tau }_k\right){\Delta }_k\nu \text{.}}$

It can be shown that, as we take partitions of [a, b] in which the points of subdivision, t_k , are chosen more and more closely together, so the corresponding sums (2.10) tend to some definite limiting value. This limit is called the Riemann-Stieltjes (RS) integral of f with respect to ν, from t = a to t = b, and we write

(2.11) ${\displaystyle {\int }_a^{b}f\left(t\right)\mathit{d\nu }\left(t\right)\equiv \underset{\mathit{\Delta t}\to 0}{lim}{\displaystyle \sum _{k=1}^{n}f\left({\tau }_k\right){\Delta }_k\nu }}$

where Δt = max(t_k –t _k–1) for 1 ≤ k ≤ n. The function ν(t) in the integral is often called the integrator.

In the particular case in which ν(t) = t, so that Δ _kν = t_k –t _k–1, the RS-integral reduces to the ordinary (Riemann) integral of f over [a, b]:

${\displaystyle {\int }_a^{b}f\left(t\right)\mathit{d\nu }\left(t\right)\equiv {\displaystyle {\int }_a^{b}f\left(t\right)\mathit{dt}=\underset{\mathit{\Delta t}\to 0}{lim}{\displaystyle \sum _{k=1}^{n}f\left({\tau }_k\right)}\left(t_k-t_{k-1}\right)\text{.}}}$

More generally, if ν is any monotone increasing function with a continuous derivative ν′, then the RS-integral of f with respect to ν can always be interpreted as an ordinary Riemann integral:

(2.12) ${\displaystyle {\int }_a^{b}f\left(t\right)\mathit{d\nu }\left(t\right)\equiv {\displaystyle {\int }_a^{b}f\left(t\right){\nu }^{\prime }\left(t\right)\mathit{dt}\text{.}}}$

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Two-Point Boundary Value Problems: Lower and Upper Solutions

Colette De Coster , Patrick Habets , in Mathematics in Science and Engineering, 2006

2.4 Bounded Solutions

In this section we consider bounded solutions of the differential equation

(2.22) $u^{″}+\mathcal{C}{u}^{\prime }=f\left(t,u\right).$

To this end, we define the set $B\mathcal{C}\left(\mathbb{R}\right)=\{u\in \mathcal{C}(\mathbb{R}\left)\right|u\in L^{\infty }\left(\mathbb{R}\right)\}$ .

Theorem 2.10. Let α and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)\cap L^{\infty }\left(\mathbb{R}\right)$ , α ≤ β and let $E:=$ $\left\{\right(t,u)\in {\mathbb{R}}^2|\alpha \left(t\right)\le u\le \beta \left(t\right)\}$ . Assume $\mathcal{C}\in \mathbb{R}$ , $F:E\to \mathbb{R}$ is a continuous bounded function and there exists M > 0 such that for all $(t,u_1)$ , $(t,u_2)\in$ E,

$u_1\le u_2\Rightarrow f\left(t,u_2\right)-f\left(t,u_1\right)\le M\left(u_2-u_1\right).$

Assume further that for all $t\in \mathbb{R}$

$\begin{array}{l}{\alpha }^{″}\left(t\right)+\mathcal{C}{\alpha }^{\prime }\left(t\right)\ge f\left(t,\alpha \left(t\right)\right),\\ {\beta }^{″}\left(t\right)+\mathcal{C}{\beta }^{\prime }\left(t\right)\le f\left(t,\beta \left(t\right)\right).\end{array}$

Let ${\alpha }_0=\alpha$ and ${\beta }_0=\beta$ . Then the equations

$\begin{array}{l}{{\alpha }^{″}}_{n+1}+c{{\alpha }^{\prime }}_{n+1}-M{\alpha }_{n+1}=f\left(t,{\alpha }_n\right)-M{\alpha }_n,\\ {{\beta }^{″}}_{n+1}+c{{\beta }^{\prime }}_{n+1}-M{\beta }_{n+1}=f\left(t,{\beta }_n\right)-M{\beta }_n\end{array}$

define sequences ${\left({\alpha }_n\right)}_n$ and ${\left({\beta }_n\right)}_n\subset B\mathcal{C}\left(\mathbb{R}\right)$ that converge monotonically and uniformly on bounded intervals of $\mathbb{R}$ to solutions u_min and u_max of (2.22) such that

$\alpha \le u_{min}\le u_{max}\le \beta .$

Further, any solution u of (2.22) with graph in E verifies

$u_{min}\le u\le u_{max}.$

Proof. First recall that given $p\in B\mathcal{C}\left(\mathbb{R}\right)$ , the problem

$y^{″}+cy-\lambda y=p\left(t\right)$

with $\lambda >0$ has a unique solution in $BC\left(\mathbb{R}\right)$ given by

$y\left(t\right)={\displaystyle {\int }_{-\infty }^{+\infty }G\left(t,s\right)p\left(s\right)}ds,$

where

$G\left(t,s\right)=-\frac{1}{2v}\exp \left(\frac{c}{2}\left(s-t\right)\right)\exp \left(-v\left|t-s\right|\right),$

with $\nu =\sqrt{\lambda +\frac{{\mathcal{C}}^2}{4}}$ . This result implies the (α_n ) _n and (β_n ) _n are uniquely defined.

Claim 1 -If α_n is such that for all $t\in \mathbb{R}$

$\begin{array}{l}{{\alpha }^{″}}_n\left(t\right)+c{{\alpha }^{\prime }}_n\left(t\right)\ge f\left(t,{\alpha }_n\left(t\right)\right),\\ {\alpha }_n\left(t\right)\le \beta \left(t\right),\end{array}$

then the function ${\alpha }_{n+1}\in B\mathcal{C}\left(\mathbb{R}\right)$ defined by

${{\alpha }^{″}}_{n+1}+c{{\alpha }^{\prime }}_{n+1}-M{\alpha }_{n+1}=f\left(t,{\alpha }_n\right)-M{\alpha }_n,$

satisffies for all $t\in \mathbb{R}$ ,

$\begin{array}{l}{{\alpha }^{″}}_{n+1}\left(t\right)+c{{\alpha }^{\prime }}_{n+1}\left(t\right)\ge f\left(t,{\alpha }_{n+1}\left(t\right)\right),\\ {\alpha }_n\left(t\right)\le {\alpha }_{n+1}\left(t\right)\le \beta \left(t\right).\end{array}$

It is enough to observe that ${\alpha }_n$ and β satisfy for all $t\in \mathbb{R}$

$\begin{array}{l}{{\alpha }^{″}}_n\left(t\right)+c{{\alpha }^{\prime }}_n\left(t\right)-M{\alpha }_n\left(t\right)\ge f\left(t,{\alpha }_n\left(t\right)\right)-M{\alpha }_n\left(t\right),\\ {\beta }^{″}\left(t\right)+c{\beta }^{\prime }\left(t\right)-M\beta \left(t\right)\le f\left(t,\beta \left(t\right)\right)-M\beta \left(t\right)\le f\left(t,{\alpha }_n\left(t\right)\right)-M{\alpha }_n\left(t\right),\end{array}$

Hence, by Theorem II-5.6, the solution ${\alpha }_{n+1}$ of

$u^{″}+c{u}^{\prime }-Mu=f\left(t,{\alpha }_n\right)-M{\alpha }_n$

is such that ${\alpha }_n\le {\alpha }_{n+1}\le \beta$ on $\mathbb{R}$ and then satisfies also for all $t\in \mathbb{R}$

$\begin{array}{l}{{\alpha }^{″}}_{n+1}\left(t\right)+c{{\alpha }^{\prime }}_{n+1}\left(t\right)-M{\alpha }_{n+1}\left(t\right)=f\left(t,{\alpha }_n\left(t\right)\right)-M{\alpha }_n\left(t\right)\\ \ge f\left(t,{\alpha }_{n+1}\left(t\right)\right)-M{\alpha }_{n+1}\left(t\right).\end{array}$

Claim 2-If ${\beta }_n$ is such that for all $t\in \mathbb{R}$

$\begin{array}{l}{{\beta }^{″}}_n\left(t\right)+c{{\beta }^{\prime }}_n\left(t\right)\le f\left(t,{\beta }_n\left(t\right)\right),\\ \alpha \left(t\right)\le {\beta }_n\left(t\right)\end{array}$

then the function ${\beta }_{n+1}$ defined by

${{\beta }^{″}}_{n+1}+c{{\beta }^{\prime }}_{n+1}-M{\beta }_{n+1}=f\left(t,{\beta }_n\right)-M{\beta }_n,$

satisffies for all $t\in \mathbb{R}$ ,

$\begin{array}{l}{{\beta }^{″}}_{n+1}\left(t\right)+c{{\beta }^{\prime }}_{n+1}\left(t\right)\le f\left(t,{\beta }_{n+1}\left(t\right)\right),\\ \alpha \left(t\right)\le {\beta }_{n+1}\left(t\right)\le {\beta }_n\left(t\right).\end{array}$

The proof of this claim is similar to the proof of Claim 1.

Conclusion - For every bounded interval $I\subset \mathbb{R}$ , we deduce from Proposition I-4.4, the existence of K such that for all n, $\Vert {\alpha }_n{\Vert }_{{\mathcal{C}}^1\left(I\right)}\le K$ and $\Vert {\beta }_n{\Vert }_{{\mathcal{C}}^1\left(I\right)}\le K$ . Hence, we deduce from the Arzelà-Ascoli Theorem the convergence to solutions u_min and u_max . As usually if u is a solution such that $\alpha \le u\le \beta$ , we can take u as a lower solution and prove that $u_{\max }={\lim }_{n\to \infty }{\beta }_n\ge u$ . Similarly we have $u_{\min }\le u$ .

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Appendix on elliptic boundary value problems in Sobolev and Hölder spaces

Ognyan Kounchev , in Multivariate Polysplines, 2001

Theorem 23.1

Let $M$ be a compact smooth manifold or $M={ℝ}^n$ . Then the following properties hold:

•

The space H^s (ℝ ⁿ ) is Hilbert with inner product

${〈u,v〉}_{H^s}:={\displaystyle {\int }_{{ℝ}^n}{\left(1+{\left|\xi \right|}^2\right)}^s\widehat{u}\left(\xi \right)\overline{\widehat{v}\left(\xi \right)}d\xi .}$

•

The dual of H^s (ℝ ⁿ ) is isomorphic to H^−s (ℝ ⁿ ), and if $M$ is also a Riemannian manifold, the dual of $H^s\left(M\right)$ is isomorphic to $H^{-s}\left(M\right)$ .

•

Let us denote by C_b (ℝ ⁿ ) the set of bounded continuous functions in ℝ ⁿ . Then the following embeddings are continuous for α > 0:

$\begin{array}{c}{H}^{\left(n/2\right)+\alpha }\left(M\right)\subset \{\begin{array}{cc}{C}_b\left(M\right)& ifM={ℝ}^n,\hfill \\ C\left(M\right)& ifMisacompactmanifold,\end{array}\\ H^{\left(n/2\right)+k+\alpha }\left(M\right)\subset C^k\left(M\right)forall\mathrm{int}egersk\ge 0,\\ H^{\left(n/2\right)+\alpha }\left(M\right)\subset C^{\alpha }\left(M\right)forall0<\alpha <1.\end{array}$

In the last $C^{\alpha }\left(M\right)$ denotes the space of bounded functions which are Hölder continuous with exponent α (see Section 23.1.4 for the definition of the Hölder space).

•

If $M$ is compact then the embedding

$H^{s+\alpha }\left(M\right)\subset H^s\left(M\right)forall\alpha >0$

is compact.

•

The trace τf of a function f ∈ S′(ℝ ⁿ ) on the subspace $\left\{0\right\}\times {ℝ}^{n-1}\subset {ℝ}^1\times {ℝ}^{n-1}=\left\{\left(x_1,x\prime \right)\right\}$ is defined by

$\left[\tau f\right]\left(x\prime \right)=f\left(0,x\prime \right).$

For all s > 1/2 the map τ extends to a unique surjective continuous map

$\tau :H^s\left({ℝ}^n\right)\to H^{s-1/2}\left({ℝ}^{n-1}\right).$

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Advanced Theory

In Pure and Applied Mathematics, 1986

14.3.2 Lemma

If(X, μ) is a regular Borel measure space, Y is a completely regular space such that C(Y) is countably generated, and η is a mapping of X into Y that is Borel and one-to-one on a compact subset $\mathcal{K}$ , then there is a Borel subset $\mathcal{K}$ ₀ of $\mathcal{K}$ such that η is a Borel isomorphism on $\mathcal{K}$ ₀ and μ( $\mathcal{K}$ \ $\mathcal{K}$ ₀) = 0.

Proof. Let $\mathcal{F}$ ₀ be the (denumerable) subalgebra over the rationals generated by a countable generating family of functions in C(Y) (so that $\mathcal{F}$ ₀ is norm, dense in the algebra C(Y ) of continuous bounded functions on Y). Since η is a Borel mapping on $\mathcal{K}$ , f $\mathcal{C}$ η is a Borel function on $\mathcal{K}$ for each f in C(Y). Let f ₁, f ₂, … be an enumeration of the functions in $\mathcal{F}$ ₀ By Lusin'stheorem, given a positive ɛ, there is a closed subset $\mathcal{K}$ ₁ of $\mathcal{K}$ such that μ( $\mathcal{K}$ \ $\mathcal{K}$ ₁) < ɛ/2 and f ₁ $\mathcal{C}$ η is continuous on $\mathcal{K}$ ₁. Again, there is a closed subset $\mathcal{K}$ ₂ of $\mathcal{K}$ ₁ such that f ₂ $\mathcal{C}$ η is continuous on $\mathcal{K}$ ₂ and μ( $\mathcal{K}$ ₁\ $\mathcal{K}$ ₂) < ɛ/4. Inductively, there is a closed subset $\mathcal{K}$ _n of $\mathcal{K}$ _{n − 1} such that f _n $\mathcal{C}$ η is continuous on $\mathcal{K}$ _n and μ( $\mathcal{K}$ _{n − 1}\ $\mathcal{K}$ _n ) < ɛ/2 ⁿ . Let $\mathcal{K}$ _ɛ be ∩ _n=1 ^∞ $\mathcal{K}$ _n . Then μ( $\mathcal{K}$ \ $\mathcal{K}$ _ɛ) < ɛ and $\mathcal{K}$ _ɛ is a closed set on which all f _n $\mathcal{C}$ η are continuous. Since {f _n } determines the topology of Y, η is a continuous mapping of $\mathcal{K}$ _ɛ into Y. As η is one-to-one on $\mathcal{K}$ _ɛ, Y is Hausdorff, and $\mathcal{K}$ _ɛ is compact, η is a homeomorphism on $\mathcal{K}$ _ɛ. It follows that η( $\mathcal{K}$ _ɛ) is compact and that the image of each Borel set in $\mathcal{K}$ _ɛ under η is a Borel set in η( $\mathcal{K}$ _ɛ) and hence in Y.

Let $\mathcal{K}$ ₀ be ∪ _n=1 ^∞ $\mathcal{K}$ _1/n . Then $\mathcal{K}$ ₀ is a Borel subset of X, $\mathcal{K}$ ₀⊆ $\mathcal{K}$ , and μ( $\mathcal{K}$ \ $\mathcal{K}$ ₀) = 0. If X ₀ is a Borel subset of $\mathcal{K}$ ₀, then X ₀∩ $\mathcal{K}$ _1/n is a Borel subset of $\mathcal{K}$ _1/n for each n; and η(X ₀) = ∪ _n=1 ^∞η(X ₀∩ $\mathcal{K}$ _1/n ). Thus η(X ₀) is a Borel set in Y and η is a Borel isomorphism on $\mathcal{K}$ ₀.

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