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The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear differential operator on Euclidean space has a distributional fundamental solution. With $D_j={\partial }_{x_j}$ and $P(D)={\sum }_{|\alpha |\le N}{a}_{\alpha }{D}^{\alpha }$, the conclusion is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^d)$ such that $P(D)E={\delta }_0$. If $N=0$, simply take $E=a_0^{-1}{\delta }_0$; hence assume $N\ge 1$.

We construct a Hörmander staircase in frequency space. The highest-degree homogeneous part $P_N$ is not identically zero on real vectors, since a polynomial vanishing on all real vectors has all coefficients zero. After an orthogonal transformation of coordinates, we can arrange that $P_N(e_1)\ne 0$. Thus, for each real ${\xi }^{\prime }\in {\mathbb{R}}^{d-1}$, the polynomialhas degree $N$ in $z$ and the same nonzero leading coefficient $b=i^N{P}_N(e_1)$, independent of ${\xi }^{\prime }$. This constant leading coefficient of a polynomial is what makes a uniform staircase possible.

For a fixed ${\xi }^{\prime }$, factor $p(z,{\xi }^{\prime })=b{\prod }_{\nu =1}^N(z-z_{\nu })$, counting repeated roots of a polynomial. Consider the $N+1$ heights $h_j=3j$, $j=0,\dots ,N$. A given root has imaginary part within distance less than one of at most one of these heights. By the pigeonhole principle, some height avoids every root, and thenThis is finite-height polynomial root avoidance. Root labels need not be chosen continuously or even measurably. Instead define the closed setsContinuity in $s$ extends the inequality from rational $q$ to every real $s$. These Borel sets partition ${\mathbb{R}}^{d-1}$. The Hörmander staircase assigns the horizontal contour $s+i{h}_j$ over $A_j$; its height is bounded by $3N$, and the denominator has the uniform lower bound $|b|$. For $d=1$, the transverse space is a single point and only one horizontal contour is needed.

Use the Fourier transform convention $\hat{\phi }(\zeta )=\int e^{-ix\cdot \zeta }\phi (x)dx$, with inverse factor $(2\pi {)}^{-d}$. Define the candidate fundamental solution of a linear differential operator byFor a test function supported in a fixed compact set $K$, Fourier decay in a bounded complex strip gives, for any integer $M$,Indeed, this is the real Fourier transform of $e^{-h_j{x}_1}\phi (x)$ with reversed frequency, and repeated integration by parts with $(1-\Delta {)}^M$ proves the estimate. Taking $2M>d$ and using the denominator bound proves absolute convergence and a continuity estimate on ${\mathcal{D}}_K$. Thus $E$ is a distribution; no unsupported interpretation of a divergent inverse Fourier transform is being used.

For its distributional derivatives, the formal transpose of a differential operator is $P(-D)$. Sinceapplying $P(D)$ cancels the denominator. For each fixed ${\xi }^{\prime }$, the numerator is an entire function of $z$. The Cauchy integral theorem, applied to a rectangle between the lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=h_j$, givesThe two vertical edges tend to zero by the same bounded-strip decay. The resulting contour deformation is performed separately for each transverse frequency, so discontinuities of the staircase height introduce no additional boundary terms. Absolute convergence permits integration over the partition $A_j$. The Fourier inversion theorem then yieldsThis proves the Malgrange–Ehrenpreis theorem. Undoing the orthogonal change of coordinates gives the fundamental solution of a linear differential operator for the original operator; the Dirac delta distribution is unchanged by that change of coordinates.

For a one-dimensional operator $L=P(D)$ of degree $N\ge 1$, with leading coefficient $a_N$, we may choose the single staircase contour below every pole. Take $\gamma >\max \{\mathrm{Re}\lambda :P(\lambda )=0\}$ and integrate on $\mathrm{Im}\zeta =-\gamma$. The corresponding formula is the Bromwich contour version of the construction above, after putting $\lambda =i\zeta$. Its poles all lie above the frequency contour. For $x<0$, close that contour downwards; the exponential function $e^{ix\zeta }$ decays and there are no enclosed poles. For $x>0$, close upwards, where the exponential function again decays, and apply the residue theorem. The bound $1/P(i\zeta )=O(|\zeta {|}^{-N})$ justifies the large arcs, including $N=1$ by the Jordan lemma away from their endpoints. Thus away from $x=0$ the retarded fundamental solution is $H(x)u(x)$, whereThe sum is over distinct roots of a polynomial, with the residue including the whole multiplicity. For simple characteristic roots of a constant-coefficient differential equation, it reduces to $u(x)={\sum }_{\lambda }{e}^{\lambda x}/P^{\prime }(\lambda )$. Repeated roots give exponential polynomial solutions of a constant-coefficient differential equation through differentiation of $e^{zx}$.

To establish the equality also at the origin, rather than leave a possible point-supported term undecided, verify the distributional jump formula for a Heaviside product. The residue expression is an entire function of $x$, andIts initial derivatives areThese identities follow by integrating over a large circle: for $j\le N-2$ the integrand is $O(|z{|}^{-2})$, whereas for $j=N-1$ its coefficient of $z^{-1}$ is $a_N^{-1}$. The distributional jump formula for a Heaviside product therefore gives $L(Hu)={\delta }_0$. Moreover, since $e^{-\gamma x}Hu$ is a tempered distribution, its Fourier transform satisfiesThere are no real zeros of this polynomial, so its reciprocal is the transform. This is exactly the shifted-contour construction, proving equality there as a distribution as well. HenceThe sign of $a_N$ matters: for the operator in the preceding solution, $a_N=-1$ and the last initial derivative is $-1$.

For fixed $u>0$, define the left-limit sigma-algebra by ${\mathcal{F}}_{u-}=\sigma ({\bigcup }_{s<u}{\mathcal{F}}_s)$. The section at $u$ of a generating rectangle $A\times (s,t]$ of the predictable sigma-algebra is $A$ when $s<u\le t$, and empty otherwise. In the first case $A\in {\mathcal{F}}_s\subseteq {\mathcal{F}}_{u-}$. Sections of the time-zero generators are empty for $u>0$.

The collection of subsets of sample-time space whose section at $u$ belongs to ${\mathcal{F}}_{u-}$ is a sigma-algebra: sections commute with complements and countable unions. It therefore contains all of $\mathcal{P}$. Apply this to $\{(\omega ,t):H_t(\omega )\in E\}$ for every Borel set $E$. Thus predictable sections at deterministic times satisfyAt $u=0$ the same argument gives ${\mathcal{F}}_0$-measurability, using the convention ${\mathcal{F}}_{0-}={\mathcal{F}}_0$. The PDF asks for information strictly before $u$; the TeX aid drops the minus sign.

The Cameron-Martin space of a Gaussian random variable in a Banach space is most conveniently defined using the first Gaussian chaos. Work over the real separable Banach space $B$, with continuous dual space $B^{\ast }$. For a centered Gaussian random variable in a Banach space, defineThe set inside the closure is already a linear subspace. Every $g\in \mathcal{G}$ is a centered Gaussian random variable. The Bochner integral defining $S$ exists: Fernique's theorem states that for some $a>0$, $\mathbb{E}{e}^{a\Vert X{\Vert }_B^2}<\infty$, and then Cauchy-Schwarz inequality gives $\mathbb{E}\Vert Xg{\Vert }_B<\infty$.

The map $S$ is injective. Indeed, $Sg=0$ implies $\mathbb{E}[g\ell (X)]=\ell (Sg)=0$ for all $\ell \in B^{\ast }$. Density of these observations in $\mathcal{G}$ then implies $\mathbb{E}{g}^2=0$. Define the Reproducing-kernel Hilbert space byThe injectivity makes the inner product unambiguous, and the completeness of $\mathcal{G}$ makes $H$ a Hilbert space. The embedding into $B$ is continuous sinceFor $\ell \in B^{\ast }$ put $Q\ell =S(\ell (X))$. The reproducing property isThis defines the Banach-space Gaussian RKHS even for a degenerate Gaussian measure.

For $h=Sg\in H$, write $\hat{h}(X)=g$. There is a measurable coordinate $\hat{h}$ on $B$, obtained as a limit in L2 space of continuous linear observations, and $\hat{h}(X)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$. The Cameron-Martin theorem for a Gaussian measure, in its real separable Banach space form, states that if $\mu$ is the probability law of $X$, then the probability law ${\mu }_h$ of $X+h$ is an equivalent probability measure to $\mu$ precisely when $h\in H$. For such a shift,For $h\notin H$ the two laws are mutually singular measures. The coordinate $\hat{h}$ need not be a continuous functional on $B$, and the sample $X$ need not belong to $H$.

Apply this theorem to the shift $-h$. With $a=\Vert h{\Vert }_H^2/2$, it givesChoose the measurable coordinate so that $\hat{h}(-x)=-\hat{h}(x)$ on a symmetric set of full $\mu$-measure. This is possible by taking an almost-surely convergent subsequence of the approximating linear observations and intersecting its convergence set with its negative. Alternatively the joint law of $(X,\hat{h}(X))$ is invariant under simultaneous negation, directly from those observations and their mean-square convergence. Since $C=-C$ and the centered Gaussian measure is symmetric, the two exponential integrals with signs $+$ and $-$ agree. Consequentlybecause $\cosh v\ge 1$. The symmetric Gaussian translation lower bound is thereforeOnly symmetry and Borel measurability of $C$ are required; convexity is not needed.

There is one degenerate flaw in the last printed claim. The separable Banach space $B=\{0\}$ with $X=0$ has $H=\{0\}$ dense in $B$, and all positive-radius balls have probability one. Nevertheless $\Phi (t)=1$ for every $t>0$. Thus the statement needs the additional hypothesis $B\ne \{0\}$. If the course convention excludes the zero space, this hypothesis is already implicit.

Under this necessary hypothesis, density of $H$ implies that $H$ contains a nonzero vector. Given $0<a<b$, scale that vector to obtain $h\in H$ with $\Vert h{\Vert }_B=(a+b)/2$, and choose $0<\delta <(b-a)/2$. The triangle inequality givesThe centered closed ball $C_{\delta }=\{x:\Vert x{\Vert }_B\le \delta \}$ is a symmetric Borel set. Apply the symmetric Gaussian translation lower bound to getHence the Gaussian norm distribution is strictly increasing when $B$ is nonzero. This proves the strict increase of a Gaussian norm distribution without assuming that its distribution has a density or that spheres have zero probability.

Again use $D_j=-i{\partial }_j$ and distribution dual pairings with complex bilinearity. The Malgrange–Ehrenpreis theorem asserts that every nonzero constant-coefficient differential operator defined by a polynomial has a fundamental solution of a linear differential operator:If $P=a\ne 0$ is constant, take $E={\delta }_0/a$. For positive degree $m$, write $P_m$ for the highest homogeneous part. A nonzero polynomial cannot vanish on all of ${\mathbb{R}}^n$, so choose a real unit vector $e$ with $P_m(e)\ne 0$. A change of coordinates by an orthogonal matrix makes it the last coordinate direction. In these coordinates,Crucially the leading coefficient $a$ is constant and never vanishes as ${\xi }^{\prime }\in {\mathbb{R}}^{n-1}$ varies.

The finite-height polynomial root avoidance construction gives the needed Hörmander staircase. Set $h_{\ell }=3\ell$ for $0\le \ell \le m$. For each fixed ${\xi }^{\prime }$, the fundamental theorem of algebra supplies $m$ roots $z_1,\dots ,z_m$, with multiplicity. A root's imaginary part can be within distance strictly less than one of at most one candidate height, since those heights are separated by three. The pigeonhole principle therefore leaves at least one height with $|h_{\ell }-\mathrm{Im}{z}_j|\ge 1$ for all roots. At that height, for every real $s$,No continuous labelling of the roots is required. To select the heights measurably, defineEach $B_{\ell }$ is closed, since the coefficients are continuous in ${\xi }^{\prime }$. Continuity in $s$ makes the rational test equivalent to the bound for all $s\in \mathbb{R}$. The $A_{\ell }$ are therefore disjoint Borel sets covering ${\mathbb{R}}^{n-1}$. The union of horizontal fibresis a Hörmander staircase, with bounded imaginary height and a uniform nonzero denominator. For $n=1$, the transverse space has one point, and this is simply the choice of one horizontal line.

For example, $P({\xi }_1,z)={\xi }_1^2+z^2$ has roots $z=\pm i{\xi }_1$. One may select $h=0$ for $|{\xi }_1|\ge 1$, and $h=3$ for $|{\xi }_1|<1$. Then the first region has $|P|\ge 1$, while in the second each root is at least two units from the chosen height. The figure shows the height projection of these horizontal fibres, rather than additional connecting contours.

Define a linear functional on test functions byTo establish that this is a distribution, fix a compact set containing the support of $\phi$, inside a radius-$R$ ball. The Fourier decay in a bounded complex strip estimate follows by applying $(1-\Delta {)}^L$ to $e^{-h_{\ell }{x}_n}\phi (x)$:The finite bound on the heights absorbs all exponential factors and powers of $h_{\ell }$ into $C_{R,L}$. Taking $2L>n$ and using $|P|\ge |a|$ proves absolute convergence and the required finite-order continuity on each fixed support set. Hence $E\in {\mathcal{D}}^{\prime }$.

Now the formal transpose identity and the Fourier derivative identity giveThe polynomial denominator cancels in the pairing for $P(D)E$. For each fixed real ${\xi }^{\prime }$, the remaining function of $z$ is entire and rapidly decreasing in its real part throughout $0\le \mathrm{Im}z\le 3m$. The Cauchy integral theorem therefore allows each horizontal fibre to be shifted to the real axis; its truncated vertical end integrals tend to zero. Absolute convergence justifies Fubini's theorem and the finite sum over the measurable partition. ThusThe last equality is Fourier inversion. Transforming back from the orthogonal coordinates preserves ${\delta }_0$ and gives the required fundamental solution of a linear differential operator for the original operator. The staircase is allowed to jump: after cancellation the shift is performed on each one-dimensional fibre, so no assertion that $\Sigma$ is a smooth contour is used. This construction proves a distributional fundamental solution, without claiming a tempered-growth estimate.

For $f\in \mathcal{D}({\mathbb{R}}^n)$, use the smoothing convolution with a test functionIndeed $u_0(x)=⟨E_y,f(x-y)⟩$ is a smooth function, with ${\partial }^{\alpha }{u}_0(x)=⟨E_y,{\partial }^{\alpha }f(x-y)⟩$. On each compact set of $x$ values, all translated test functions have supports in one fixed compact set, so differentiation is justified by the continuity of the distribution. Compact support of $f$ makes the convolution well defined even when $E$ is not tempered. It does not imply compact support of $u_0$.

Finally the affine solution space of a linear equation isIf every homogeneous solution is represented by a smooth function, so is every $u=u_0+v$. Conversely, if every solution for this fixed $f$ is represented by a smooth function, then for any homogeneous solution $v$, the solution $u_0+v$ is smooth, and subtracting the smooth $u_0$ shows that $v$ is smooth. Hence All inhomogeneous solutions are smooth functions exactly when all homogeneous solutions are smooth functions. This is a global statement on ${\mathbb{R}}^n$ for the stated data; no additional local regularity theorem is being assumed.

Let $q$ be conjugate to $p$, so $1/q=1-1/p$, including $q=1$ for $p=\infty$. Hölder's inequality implies, for every measurable finite-measure set $E$,This proves integrability on $A$ and gives uniform integrability there. Choose a set $B\subseteq A$ with $\lambda (A\setminus B)\le \delta$ on which convergence is uniform, using Egorov theorem. ThenFirst take $n\to \infty$ and then $\delta \downarrow 0$. Thus even the local $L^1$ difference tends to zero, and in particularFor $1<p<\infty$, simple functions supported on finite-measure Borel sets are dense in $L^q(\mathbb{R})$. The displayed convergence proves pairing convergence for such functions. Approximate any $h\in L^q$ by a simple $h_0$ and useBy the duality of Lp spaces, these are all continuous linear functionals on $L^p$, so $f_n$ converges weakly to $f$. It need not converge strongly: $f_n=1_{[n,n+1]}$ converges pointwise to zero but has $L^p$ norm one. For $p=\infty$ the pairing argument gives weak-star convergence against $L^1$, rather than asserting weak convergence against every functional in the much larger full dual.

The chosen non-eventually-one binary expansion fixes the ambiguity at dyadic rationals. Its $n$th digit isIn particularThis is a finite union of Borel sets; the zero-digit preimage is its complement in $[0,1)$. Since $R_n$ only takes the values zero and one, every Borel preimage is a union of these two sets. Thus it is Borel measurable. The half-open intervals implement the terminating-zero convention at dyadic endpoints.

Use the usual unital-subalgebra convention $I_H\in A$. The spectral theorem for a commutative operator algebra states that the character space ${\Phi }_A$, equipped with the Gelfand topology, is a compact Hausdorff space, and there is a unique regular projection-valued measure $E$ on its Borel sigma-algebra satisfyingwhere $\hat{a}(\chi )=\chi (a)$ is the Gelfand transform. The map $a\mapsto \hat{a}$ is an isometric unital star-isomorphism from $A$ onto $C({\Phi }_A)$ by the Commutative Gelfand--Naimark theorem.

More explicitly, every $E(B)$ is an orthogonal projection, $E(\varnothing )=0$, $E(B\cap C)=E(B)E(C)$, and for disjoint Borel sets $B_n$,with convergence in the norm of the Hilbert space. Regularity means that each scalar spectral measure $B\mapsto ⟨E(B)v,w⟩$ is a regular Borel measure. The integral identity then implies $\Vert a\Vert ={\sup }_{\chi \in {\Phi }_A}|\hat{a}(\chi )|$. No separability assumption on $H$ is needed.

A subset of $\mathbb{R}$ is a Borel set when it belongs to the Borel sigma-algebra, the smallest sigma-algebra containing every open set of $\mathbb{R}$.

The rational numbers are countable, sois a countable union of closed sets and hence a Borel set; its Lebesgue measure is zero. Its complement is also Borel and has infinite measure:

The intervals are nested in the relevant directions, and the strongest lower and upper restrictions both occur at $n=2$:Thus this set is closed, hence Borel, and

A function $f:\mathbb{R}\to \mathbb{R}$ is Borel measurable whenis a Borel set for every Borel set $B\subseteq \mathbb{R}$. It is enough to require this for every open set, or for every interval of the form $(-\infty ,a)$.

Yes. In fact, measurability of $f$ is unnecessary. Define the oscillationFor every $\epsilon >0$, the set $\{x:{\omega }_f(x)<\epsilon \}$ is open, and $f$ is continuous at $x$ exactly when ${\omega }_f(x)=0$. Hence the continuity set of a function isa $G_{\delta }$ set and therefore a Borel set.