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Let $u_j\in C_c^{\infty }({\mathbb{R}}^n)$ tend to $u$ in the Sobolev space $W^{1,p}$, using density of smooth functions in a Sobolev space. Write $\alpha =1-n/p>0$. Applying the Morrey inequality on a cube to a unit cube centered at $x$, and bounding its average by the Holder inequality, givesfor every smooth $v$. Hence $u_j$ is a Cauchy sequence in the supremum norm and has a bounded continuous limit $u^{\ast }$ by uniform convergence. On every finite-measure cube, this limit is also the $L^p$ limit, so $u=u^{\ast }$ almost everywhere.

For $x\ne y$, choose $r=4|x-y|$, so that $y\in Q_r(x)$. The second estimate from part (a), followed by passage to the limit, givesThus $u^{\ast }$ is a Hölder continuous function andIt is the unique continuous representative: two continuous functions agreeing almost everywhere agree everywhere.

For $1\le p<n$, take a smooth cutoff function $\chi$ equal to one near zero and setIts weak derivative has size $O(|x{|}^{-a-1})$ near zero, whose $p$th power is integrable because $p(a+1)<n$. Thus $u\in W^{1,p}$, but it is essentially unbounded in every neighborhood of zero and has no continuous representative. The value assigned at zero is irrelevant.

The Lax-Milgram theorem assumes that $H$ is a real Hilbert space and $B:H\times H\to \mathbb{R}$ is a bounded bilinear form and a coercive bilinear form: there are $M,\alpha >0$ such thatThen for every continuous linear functional $\ell \in H^{\prime }$ there is a unique $u\in H$ satisfyingSymmetry of $B$ is not required.

For the proof, the Riesz representation theorem provides a bounded linear operator $T:H\to H$ and $g\in H$ with $B[u,v]=(Tu,v{)}_H$ and $\ell (v)=(g,v{)}_H$. Coercivity and the Cauchy-Schwarz inequality implyThus $T$ is injective and its range is closed: if $T{u}_j$ converges, the displayed inequality makes $u_j$ a Cauchy sequence; completeness gives $u_j\to u$ and $T{u}_j\to Tu$. If $w$ lies in the orthogonal complement of the range, then $B[v,w]=0$ for all $v$. Taking $v=w$ gives $\alpha \Vert w{\Vert }^2\le B[w,w]=0$, so $w=0$. A closed subspace with zero orthogonal complement is all of $H$, hence $T$ is surjective. The unique solution is $T^{-1}g$, and the same inequality gives the asserted norm bound.

Set $M_n={\sum }_{j=0}^n{X}_j$ and ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$. The independence and zero expected values give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$, so $(M_n)$ is a martingale. The variance additivity for independent random variables givesThe Cauchy-Schwarz inequality implies ${\sup }_n\mathbb{E}|M_n|\le \sqrt{V}$. Applying the Martingale convergence theorem from (b),In fact the L2 martingale convergence theorem also gives convergence in L2. Independently, for $n>m$ the same variance additivity for independent random variables yields $\mathbb{E}|M_n-M_m{|}^2={\sum }_{j=m+1}^n\mathbb{E}[X_j^2]\to 0$, confirming that the partial sums form a Cauchy sequence in $L^2$.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

The Moore–Penrose inverse of an operator is the linear mapthat assigns each admissible datum its minimum-norm least-squares solution. If $f=y+z$ with $y\in \mathcal{R}(K)$ and $z\in \mathcal{R}(K{)}^{\perp }$, then $K^{†}f$ is the unique $u\in \mathcal{N}(K{)}^{\perp }$ satisfying $Ku=y$. In particular it vanishes on $\mathcal{R}(K{)}^{\perp }$, andThe continuity criterion, with the inherited data norm on its domain, isFor a closed operator range, the restriction of $K$ from $\mathcal{N}(K{)}^{\perp }$ to $\mathcal{R}(K)$ is a bounded bijection between Banach spaces, so its inverse is bounded by the bounded inverse theorem. Conversely, if $K^{†}$ is bounded and $K{u}_n\to y$, replace each $u_n$ by $P_{\mathcal{N}(K{)}^{\perp }}{u}_n=K^{†}K{u}_n$. This is a Cauchy sequence, hence converges to $u$, and $Ku=y$. Thus every limit of range elements remains in the operator range.

A metric space is complete when every Cauchy sequence converges to a point of the space.

Let $x^{(1)},x^{(2)},\dots$ be a Cauchy sequence in $X$, where $x^{(j)}=(x_1^{(j)},x_2^{(j)},\dots )$. For each positive integer $N$, choose $J_N$ such thatThis inequality says that the first $N$ coordinates of $x^{(j)}$ and $x^{(k)}$ agree. Consequently each coordinate $x_n^{(j)}$ is eventually constant. Let $x_n$ be its eventual value and put $x=(x_1,x_2,\dots )\in X$.

For $j\ge J_N$, the first $N$ coordinates of $x^{(j)}$ agree with those of $x$, soAs $N$ is arbitrary, $x^{(j)}\to x$. Therefore

SupposeIf $(v_n)$ is a Cauchy sequence in $\Vert \cdot \Vert$, thenConversely, $\Vert v_n-v_m\Vert \le c^{-1}\Vert v_n-v_m{\Vert }^{\prime }\to 0$. Hence the sequence is Cauchy for one norm exactly when it is Cauchy for the other.

First, compactness makes $X$ totally bounded: for every positive integer $k$, the open cover by balls of radius $1/k$ has a finite subcover. Given a sequence, repeatedly choose a ball of radius $1/k$ containing infinitely many terms of the subsequence retained at the preceding stage. Taking a diagonal subsequence $(x_{n_k})$, any two terms with indices at least $k$ lie in one ball of radius $1/k$, so their distance is below $2/k$. The diagonal subsequence is therefore a Cauchy sequence.

A compact metric space is complete. To see this without assuming sequential compactness, let $(y_n)$ be Cauchy and put $F_n=\overline{\{y_m:m\ge n\}}$. These nonempty closed sets are nested and have the finite intersection property; compactness gives a point $y\in {\bigcap }_n{F}_n$. Given $\epsilon >0$, a sufficiently late tail has diameter below $\epsilon /2$, and because $y$ lies in its closure, every point of that tail lies within $\epsilon$ of $y$. Thus $y_n\to y$.

The Cauchy subsequence constructed above consequently converges in $X$. Hence every compact metric space is sequentially compact.

Let $(x_n)$ be a Cauchy sequence in a compact metric space $X$. By sequential compactness of a compact metric space, it has a convergent subsequence $x_{n_j}\to x\in X$. Given $\epsilon >0$, choose $N$ such that $d(x_m,x_n)<\epsilon /2$ whenever $m,n\ge N$, and then choose $j$ such that $n_j\ge N$ and $d(x_{n_j},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole sequence converges to $x$. Every Cauchy sequence converges in $X$, so $X$ is a complete metric space.

Given $x\in {\ell }^p$, choose a sequence $x^{(m)}\in F$ with $x^{(m)}\to x$, which is possible by part (a). The bound givesso $(T{x}^{(m)})$ is a Cauchy sequence. Since ${\ell }^p$ is a Banach space, it converges. DefineThe same estimate shows that this limit does not depend on the approximating sequence. Passing to the limit proves linearity andso $\Vert \tilde{T}\Vert \le C$. It agrees with $T$ on $F$. If two bounded extensions existed, their difference would vanish on the dense subspace $F$ and, by continuity, on its closure ${\ell }^p$. Thus the extension is unique. This is the extension of a bounded linear operator from a dense subspace.

The Baire category theorem says that a complete metric space cannot be a countable union of nowhere dense sets, equivalently that every countable intersection of open dense sets is dense. To prove it, let $G_1,G_2,\dots$ be open dense subsets and let $V$ be nonempty and open. Choose nested closed ballsstarting inside $V\cap G_1$. The centres are a Cauchy sequence; completeness gives a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense.

Choose a countable sequence $q_j\uparrow p$ with $1\le q_j<p$. Since ${\ell }^q\subset {\ell }^r$ for $q<r$, every ${\ell }^q$ with $q<p$ lies in some ${\ell }^{q_j}$. For integers $m\ge 1$, setPointwise convergence and Fatou lemma for series show that $E_{j,m}$ is closed in ${\ell }^p$. It has empty interior: inside any ${\ell }^p$ ball, add a sufficiently long finite block whose ${\ell }^p$ norm is tiny but whose ${\ell }^{q_j}$ norm is larger than $m$. Therefore every $E_{j,m}$ is nowhere dense. Since ${\ell }^p$ is a Banach space, Baire gives

An explicit element of the difference isIts $p$th powers form a convergent logarithmic series, whereas for every $q<p$ the powers contain $n^{-q/p}$ with exponent below one and their series diverges. Hence $x\in {\ell }^p$ but $x\notin {\ell }^q$ for every $q<p$.

For $n>m$, independence and $\mathbb{E}{X}_k=0$ giveThus $(S_n)$ is a Cauchy sequence in the complete L2 space and converges in $L^2$ to some random variable $S$. The Cauchy-Schwarz inequality gives $\Vert Z{\Vert }_1\le \Vert Z{\Vert }_2$ on a probability space, so the convergence also holds in $L^1$. Finally, convergence in $L^2$ implies convergence in probability, which implies convergence in distribution.

One form of the Hensel lemma is the following. Let $R$ be a complete discrete valuation ring with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $a_0\in R$ satisfiesthen there is a unique $a\in R$ with $f(a)=0$ and $a\equiv a_0(\mathrm{mod}\mathfrak{m})$.

Define $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Since $f^{\prime }(a_n)$ remains a unit, Taylor expansion givesso the valuations of the errors at least double. The corrections tend to zero, making $(a_n)$ a Cauchy sequence; completeness gives a limit $a$, and continuity gives $f(a)=0$. If $a,b$ are two such roots, thenwith $u\in \mathfrak{m}$, so the second factor is a unit and $a=b$.

For $T_2>T_1\ge -1$, the dual Strichartz estimate for the free Schrödinger equation givesThe assumed finite global $L_{t,x}^4$ norm makes the right side tend to zero as $T_1,T_2\to \infty$. The displayed family is therefore a Cauchy sequence in the complete space $L^2$ and has a strong limit $F_{\infty }$.

The Duhamel principle givesDefineSince the free Schrödinger group is unitary,This is scattering from a finite Strichartz norm.

A Cauchy sequence is a sequence $(x_n)$ such that for every $\epsilon >0$ there is $N$ for whichThe general principle of convergence, or completeness of the real numbers, states that a real sequence converges if and only if it is Cauchy.

For the final claim, let $m=\lfloor n/2\rfloor$. Since $(a_n)$ is decreasing and positive,Because the series $\sum a_k$ converges, its tails tend to zero. Also $n-m\ge n/2$, soThe squeeze theorem gives

This is always true. Uniform convergence makes $(f_n)$ a uniformly Cauchy sequence: for every $\epsilon >0$, there is $N$ such thatfor every $x\in A$ and $m,n\ge N$. Letting $x\to x_0$ through points of $A\setminus \{x_0\}$ and using the assumed limits givesThus $(y_n)$ is a Cauchy sequence. Since $Y$ is a complete metric space, $(y_n)$ converges in $Y$.

Choose a test function $\chi$ that equals one on the unit ball and vanishes outside the ball of radius two. Sinceintegration by parts givesThe first term on the right is supported where $1\le |x|\le 2$, where $u^2\le |x{|}^{\alpha }{u}^2$. The Cauchy-Schwarz inequality and Young inequality bound the second term byAfter absorbing the first integral,

This estimate also proves completeness. Indeed, a Cauchy sequence in $\Sigma$ is Cauchy in $H^1$ locally, while its gradients and the functions $|x{|}^{\alpha /2}{u}_n$ converge in $L^2$. The limits agree locally with a function $u$, so $u\in \Sigma$ and $u_n\to u$ in the energy norm. Thus $\Sigma$ is a Hilbert space; it is the confining-potential energy space for $V(x)=|x{|}^{\alpha }$.

Let $(u_n)$ be a Cauchy sequence in the graph normThen $(u_n)$ and $(A{u}_n)$ are Cauchy sequences in the Banach space $H$, so for some $u,v\in H$,Because $A$ is a closed linear operator, $u\in D(A)$ and $Au=v$. It follows thatThus every Cauchy sequence converges in $Y=(D(A),\Vert \cdot {\Vert }_Y)$, and