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Write $y={\psi }_s$ and retain the Landweber relaxation parameter $\tau$. The printed $\Vert A\Vert <2$ allows the fixed choice $\tau =1/2$, since then $\tau \Vert A{\Vert }^2<2$. It does not justify a unit step: the scalar linear operator $A=3/2$ has norm below $2$, but unit-step error is multiplied by $1-|A{|}^2=-5/4$ and diverges. We use $0<\tau <2/\Vert A{\Vert }^2$ throughout.

From the recurrence and $f_0=0$, induction yields the closed formFor the compact operator in part ii, take a singular system of a compact operator with $A{v}_j={\sigma }_j{u}_j$, $A^{\ast }{u}_j={\sigma }_j{v}_j$ and ${\sigma }_j>0$. Thus ${\sigma }_j^2$ are the eigenvalues of $A^{\ast }A$; this fixes the notation implicit in the printed hint. In the singular component $j$, the recurrence isSumming the geometric series gives the Landweber spectral filterThere is no contribution from data in $\ker A^{\ast }$ or from $\ker A$ in the reconstruction. Starting at zero is what selects the minimum-norm least-squares solution.

For fixed ${\sigma }_j>0$, $|1-\tau {\sigma }_j^2|<1$, so the numerator tends to $1$. The assumption $y\in D(A^{†})$ is the Picard criterionwith an arbitrary additional component in $\ker A^{\ast }$. The squared reconstruction error isEach term tends to zero and is bounded by a summable term from the Picard criterion. The dominated convergence theorem therefore proves $R_{n}y\to A^{†}y$, where $A^{†}$ is the Moore–Penrose inverse of an operator.

For each finite $n$, the regularization of an inverse problem is stable. Set $t=\tau {\sigma }^2\in (0,2)$. The geometric series and $|1-t|\le 1$ implyConsequently the Landweber noise amplification bound in this general step-size convention is $\Vert R_n\Vert \le \sqrt{2\tau n}$. If additionally $\tau \Vert A{\Vert }^2\le 1$, the sharper numerator bound $\min (nt,1)$ gives $\Vert R_n\Vert \le \sqrt{\tau n}$.

To see the regularization parameter directly, put $\alpha =1/(\tau n)$. Modes with ${\sigma }^2\ll \alpha$ have numerator approximately $n\tau {\sigma }^2$, so their inverse coefficient is approximately $\sigma /\alpha$ rather than $1/\sigma$. Each fixed nonzero mode is eventually restored as $\alpha \downarrow 0$. ThusThe finite iterates suppress unstable small singular values; infinitely many iterations remove this suppression. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, the noise-bias decomposition for linear regularization givesChoosing $n(\delta )\to \infty$ and ${\delta }^{2}n(\delta )\to 0$ proves noisy-data convergence. The reciprocal iteration index is a regularization parameter, and early stopping controls noise amplification.

The Schwartz space consists of the smooth functions $\phi$ for whichfor all multi-indices $\alpha ,\beta$. Its topology is generated by these seminorms; ${\phi }_j\to \phi$ means convergence in every seminorm. A tempered distribution is a continuous linear functional on this space. Convergence in ${\mathcal{S}}^{\prime }$ here means weak convergence: $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every Schwartz function.

Use the unnormalized angular-frequency Fourier transformThe transform is a continuous automorphism of the Schwartz space, so this duality defines a tempered distribution. Put $D_j=-i{\partial }_j$, as required by the stated identities. Integration by parts and differentiation under the integral giveThus, keeping the transpose factor of $D=-i\partial$ explicit,andIterating gives the Fourier derivative identities for distributions

The logarithm defines a tempered distribution because it is locally integrable and grows more slowly than a polynomial. Its derivative is $u^{\prime }(x)=x/(1+x^2)$. The standard transform pair $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\lambda |}$, obtainable by the residue theorem, givesAs $\widehat{u^{\prime }}=i\lambda \hat{u}$, we have $\lambda \hat{u}=-\pi \mathrm{sgn}(\lambda )e^{-|\lambda |}$. DefineThis integral is absolutely convergent: the numerator is $O(|\lambda |)$ near zero, and the exponential controls the tails. It defines a tempered distribution satisfying the same multiplication equation. The kernel of multiplication by a coordinate therefore gives $\hat{u}=T+c{\delta }_0$.

To determine $c$, use the supplied Gaussian Fourier transform pair with ${\phi }_n(0)=1/(2\sqrt{\pi })$. ThenAlso $⟨T,{\phi }_n⟩\to 0$ by the dominated convergence theorem: the relevant integrand is bounded by a constant times $|\lambda |e^{-|\lambda |}$ for $n\ge 1$. ConsequentlyThe cutoff limit agrees with the absolutely convergent integral; this particular subtraction convention fixes the otherwise ambiguous delta term. The reusable identity is Fourier transform of the logarithm of one plus x squared.

Add and subtract the exact-data iterate. The noise-bias decomposition for linear regularization givesso the supplied Landweber noise amplification bound yieldsWith unit step and zero initial iterate, the exact-data error is $x^{†}-x_n=(I-A^{\ast }A{)}^n{x}^{†}$. Its squared norm isby the dominated convergence theorem. Stopping with $n\to \infty$ but $n{\delta }^2\to 0$ therefore makes both errors vanish.

The integrands are absolutely integrable because $f$ has compact support, $g\in L^1(\mathbb{R})$, and $|{\phi }_Z|\le 1$. Fubini's theorem therefore permits us to insert ${\phi }_Z(s)=\mathbb{E}[e^{isZ}]$ and interchange the expectation and integrals:With $r=\epsilon s$, the inner integral in $s$ isNow set $t=(Z-x)/\epsilon$. The change of variables $x=Z-\epsilon t$ giveswhich is the required identity.

Choose any $g$ with $g,\hat{g}\in L^1$, $g(0)\ne 0$, and for which Fourier inversion theorem applies, for example a Gaussian function. The left side factors through the Fourier transform of $f$ asOn the right side, $f(Z-\epsilon t)\to f(Z)$ andThe dominated convergence theorem and Fourier inversion at zero giveConsequently the Fourier-cutoff recovery of expectations from a characteristic function isThe right side depends only on ${\phi }_Z$. Equal characteristic functions therefore give equal expectations for every compactly supported continuous $f$, and part a yields the uniqueness theorem for characteristic functions.

For the heat kerneldirect partial differentiation givesThus $K$ solves the heat equation. For bounded continuous $f$, differentiation under the integral sign givesWith $Y=(x-y)/\sqrt{4t}$,The Gaussian integral makes the weight have total mass one, and the dominated convergence theorem gives $u(x,t)\to f(x)$. This is the Gaussian approximate identity.

For the viscous Burgers equation with unit viscosity, the Cole-Hopf transformation $w=-2u_x/u$ reduces the equation to $u_t=u_{xx}$. To obtain initial value $g$, choose(up to an irrelevant positive constant), and set $u=K_t\ast f$. ThereforeEquivalently,and the approximate-identity limit gives $w(x,t)\to g(x)$.

It is sufficient thatand that the stopping rule obeyFor example, $k(\delta )=\lfloor {\delta }^{-1}\rfloor$ works.

For exact data $f\in \mathcal{D}(A^{†})$, each factor $(1-\tau {\sigma }_i^2{)}^k$ tends to zero. The Picard criterion makes $⟨f,y_i⟩/{\sigma }_i$ square summable, while $|1-(1-\tau {\sigma }_i^2{)}^k|$ is uniformly bounded. The dominated convergence theorem on the resulting series yields $u_k(f)\to A^{†}f$.

For noisy data with $\Vert f^{\delta }-f\Vert \le \delta$, the filter representation givesIndeed, when $\tau {\sigma }^2\le 1$, Bernoulli's inequality gives $1-(1-\tau {\sigma }^2{)}^k\le \min (k\tau {\sigma }^2,1)$; when $\tau {\sigma }^2>1$, the quotient is uniformly bounded because $\tau \Vert A{\Vert }^2<2$. Hence the triangle inequality giveswhich proves that early-stopped Landweber iteration is a convergent regularization of an inverse problem.

The following four assumptions are sufficient, with statements understood for ${\mu }_0$-almost every $u$ and every $y$:

The first two assumptions give $0<Z(y)<\infty$. The last two allow the dominated convergence theorem to prove $L^1({\mu }_0)$ continuity of the normalized posterior density, which is equivalent to continuity in total variation distance.

For a nonnegative measurable $f$, defineEach $f_n$ takes only finitely many values, its level sets are measurable, and $f_n(x)\to f(x)$ pointwise. Thus measurable nonnegative functions are pointwise limits of simple functions.

If $f$ is integrable, the same functions satisfy $0\le f_n\le f$ and converge pointwise. The dominated convergence theorem givesThey are integrable because they are bounded above by $f$.

If $0\le f\le M$, instead setThis has finitely many values andeverywhere, so $g_n\to f$ uniformly.

The union bound gives $\mathbb{P}({\bigcup }_{n\ge m}{A}_n)\le {\sum }_{n\ge m}\mathbb{P}(A_n)\to 0$. Taking the decreasing intersection proves the first Borel-Cantelli lemma.

If $X_n\to X$ almost surely, then $1_{\{|X_n-X|>\epsilon \}}\to 0$ almost surely; dominated convergence theorem gives convergence of its expectation, hence convergence in probability.

If convergence in probability holds, any subsequence has a further one with $\mathbb{P}(|X_{n(k_r)}-X|>2^{-r})<2^{-r}$. Borel-Cantelli then gives almost-sure convergence. Conversely, failure in probability supplies a subsequence with probabilities bounded below by some positive constant, and no further subsequence can converge almost surely because that would imply convergence in probability.

After replacing $X_j$ by $X_j-\mu$, the maximal inequality for independent averages givesThis follows from the Doob Lp maximal inequality by dyadically grouping the partial sums. The strong law makesalmost surely. The displayed maximal function is in $L^p$, so dominated convergence applied to its $p$th power proves convergence in $L^p$.

Use the conventionThe Riemann-Lebesgue lemma states that if $f\in L^1({\mathbb{R}}^n)$, then $\hat{f}$ is continuous andContinuity follows from the dominated convergence theorem, since $e^{-ix\cdot {\xi }_j}\to e^{-ix\cdot \xi }$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1({\mathbb{R}}^n)$. Choose a coordinate $j$ for which $|{\xi }_j|\ge |\xi |/\sqrt{n}$. Integration by parts givesand henceSince $C_c^1({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$, choose $g$ with $\Vert f-g{\Vert }_1<\epsilon$. The elementary Fourier bound givesso the decay for $g$ implies the decay for $f$.

The Parseval identity says that for $f,g\in L^2({\mathbb{R}}^n)$, with their Fourier transforms defined by the Plancherel theorem,In particular,

For the given radial function, as $|x|\to 0$,while as $|x|\to \infty$,Using polar coordinates, local integrability of $|f{|}^q$ is therefore determined byand integrability at infinity byThe assumptions $2a>-n$ and $b>n$ imply both $f\in L^2({\mathbb{R}}^n)$ and $f\in L^1({\mathbb{R}}^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\hat{f}|\le \Vert f{\Vert }_1$ give $\hat{f}\in L^{\infty }$, while Parseval gives $\hat{f}\in L^2$. Finally, for every $2\le p<\infty$,Thus

Because the centered normal distribution density $g_t$ is nonnegative and has integral one, Fubini's theorem and translation invariance giveThus $f\ast g_t\in L^1$.

Use the angular-frequency conventionThe convolution theorem and the Gaussian transform giveSince $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:This proves the Fourier inversion theorem for the convolution directly.

Now additionally suppose $\hat{f}\in L^1$. For every $x$, the dominated convergence theorem givesFourier inversion identifies the right-hand side with $f(x)$ for Lebesgue almost everywhere $x$. Therefore the Gaussian approximate identity satisfies

Mazur theorem says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number $\alpha$ such thatThis weakly open separation shows that $x$ is outside the weak closure.

By definition,For each $n$, regard $x_n$ as the bounded functional $J{x}_n:X^{\ast }\to \mathbb{F}$ given by $J{x}_n(f)=f(x_n)$. Pointwise convergence makes the family $(J{x}_n)$ pointwise bounded on the Banach space $X^{\ast }$. The Uniform boundedness principle gives

Suppose first that $x_n\rightharpoonup 0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\epsilon$.

Conversely, if weak convergence fails, some $f\in X^{\ast }$ and an infinite subsequence satisfy either $\mathrm{Re}f(x_n)\ge \epsilon$ throughout or $\mathrm{Re}f(x_n)\le -\epsilon$ throughout. Every convex combination from that subsequence then has norm at least $\epsilon /\Vert f\Vert$, contradicting the stated property.

Now let the $x_n\in {\ell }^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\Vert x_n{\Vert }_p\le C$. For $N$ distinct terms,The convex-combination criterion therefore proves $x_n\rightharpoonup 0$. This is weak convergence of bounded disjointly supported sequences in lp.

The final implication for a commutative unital C*-algebra is true. By the Commutative Gelfand--Naimark theorem, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the Riesz-Markov-Kakutani representation theorem; the dominated convergence theorem givesHence $a_n\rightharpoonup 0$.

The martingale convergence theorem gives an almost-sure limit $M_{\infty }$, because $|M_n|\le 1$. The dominated convergence theorem also gives $M_n\to M_{\infty }$ in $L^1$.

To identify the limit, take $A\in {\bigcup }_n{\mathcal{F}}_n$. For some $N$, $A\in {\mathcal{F}}_N$, and for every $n\ge N$,Passing to the $L^1$ limit preserves this equality. The sets for which $\mathbb{E}[M_{\infty }{1}_A]=\mathbb{E}[Z{1}_A]$ form a monotone class containing the algebra ${\bigcup }_n{\mathcal{F}}_n$, so the equality holds throughout ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $M_{\infty }$ is ${\mathcal{F}}_{\infty }$-measurable, it is $\mathbb{E}[Z\mid {\mathcal{F}}_{\infty }]$. This proves the conditional-expectation convergence along a filtration both almost surely and in $L^1$.

Define $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $0\le Z_n\le 2$ and $Z_n\downarrow 0$ almost surely. The dominated convergence theorem gives $\mathbb{E}{Z}_n\to 0$.

Set $W_n=\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$. Since $Z_{n+1}\le Z_n$, the tower property of conditional expectation givesso $(W_n)$ is a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives $W_n\to W_{\infty }$ almost surely, and Fatou lemma yields $\mathbb{E}{W}_{\infty }\le {\mathrm{liminf}}_n\mathbb{E}{W}_n=0$. Hence $W_n\to 0$ almost surely; because $\mathbb{E}{W}_n=\mathbb{E}{Z}_n\to 0$, convergence also holds in $L^1$.

Finally,The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the moving-variable conditional-expectation convergence.

The almost sure supermartingale convergence theorem says that a supermartingale whose negative parts have uniformly bounded expectations converges almost surely to a finite integrable limit. In particular, every nonnegative supermartingale converges almost surely.

Here the process is uniformly bounded, say $0\le X_n\le C$. Let $X_{\infty }={\lim }_n{X}_n$, whose existence follows from the theorem. The dominated convergence theorem then givesThe limit on the right also exists directly because the expectations of a supermartingale form a decreasing sequence.

Set $g_{ϵ}(x)=x/\sqrt{{ϵ}^2+x^2}$ and $g(x)=\mathrm{sgn}(x)$, with $g(0)=0$. For every $s>0$, the normal distribution of $B_s$ has no atom at zero, so $g_{ϵ}(B_s)\to g(B_s)$ almost surely. Since $|g_{ϵ}-g|\le 2$, the dominated convergence theorem and Tonelli theorem giveThe Itô isometry followed by the Doob L2 maximal inequality now yieldsThus the stochastic integrals converge ucp to ${\int }_0^t\mathrm{sgn}(B_s)d{B}_s$.

For Brownian motion started at $x\in D$, the Itô formula and $\Delta u=0$ show thatis a local martingale. Since $D$ is bounded and $u$ is continuous on the compact set $\overline{D}$, the stopped process is bounded and hence a true martingale. Brownian motion exits every bounded domain almost surely, so $t\wedge \tau \to \tau$. The dominated convergence theorem, continuity at the boundary, and $u=f$ on $\partial D$ give the Brownian representation of the Dirichlet problem

Let $\sigma$ be the first time Brownian motion started outside the unit half-disc reaches its semicircular boundary. A path that reaches $A$ must first cross that semicircle. The Strong Markov property at $\sigma$ givesTake $z=iy$, multiply by $y$, and let $y\to \infty$. The Brownian representation of half-plane capacity identifies the left side with $\mathrm{hcap}(A)$, while part b givesSince the exit height is between zero and one, the dominated convergence theorem applies and yields

Let $T$ be a measure-preserving transformation of $(X,\mathcal{F},\mu )$ and let $f\in L^1(\mu )$. The Birkhoff ergodic theorem states thatalmost everywhere, where $\mathcal{I}$ is the invariant sigma-algebra. The limit is $T$-invariant and

Assume now that $\mu (X)=1$. For $M>0$, letbe the bounded truncation of $f$, and let $f_M^{\ast }$ be its Birkhoff limit. Since $|A_n{f}_M-f_M^{\ast }|\le 2M$ on a finite measure space, the dominated convergence theorem givesMeasure preservation and the triangle inequality give the $L^1$ contractionMoreover, the almost-everywhere convergence and Fatou lemma implyConsequentlySince $f_M\to f$ in $L^1$, letting $M\to \infty$ proves the L1 convergence in the Birkhoff ergodic theorem on a finite measure space: