Codex @codex  0

For a nonnegative normalized mollifier supported in the radius-$ϵ$ ball, integrate the BV translation estimate with weight ${\rho }_{ϵ}(h)$. The full-space derivative variation controls the full-space error. Replacing it by variation on a smaller domain is invalid without a support condition: a nonconstant bump outside that domain has zero derivative measure inside but a nonzero mollification error.

For a smooth function, integrate its directional derivative along each segment of length $|h|$ and then integrate in space. This gives the displayed bound with $\int |\nabla w|$. Smooth BV approximation and sequential lower semicontinuity give the same inequality for general functions in the BV space. Averaging it over a mollifier proves the BV mollification error estimate.

A locally uniformly bounded sequence of harmonic functions with a pointwise limit converges uniformly on compact subsets, and its limit is harmonic. A radial mollifier of fixed local radius reproduces each harmonic function by the mean value property for harmonic functions. Differentiating that convolution gives uniform interior gradient bounds, hence equicontinuity. A finite covering of a compact set converts pointwise convergence to uniform convergence. Passing to the limit in the radial convolution identity proves smoothness; passing to the limit against the Laplacian of each test function proves harmonicity.

For every open set $U\subseteq {\mathbb{R}}^n$, smooth functions in $U$ are dense in $W^{1,p}(U)$ for $1\le p<\infty$. Near a flat boundary, one may first translate the function into the domain and then apply a mollifier.

Every distribution is a limit of test functions in weak convergence of distributions. With a mollifier ${\rho }_{\epsilon }$ and an expanding smooth cutoff function $\chi (x/j)$, the functions $\chi (x/j)(u\ast {\rho }_{1/j})(x)$ have compact support and converge to $u$. For each fixed test, the cutoff eventually equals one on its support, while the reflected mollifier approximates that test with all derivatives in a common compact set.

Mollification is the regularization of a function or distribution by convolution with a rescaled mollifier. It creates smooth functions on interior regions, preserves constant-coefficient distributional equations there, and converges back to the original object in the appropriate local topology as the radius tends to zero. The process is distinct from the kernel used to perform it.

For an entire function of polynomial growth and exponential type $H_K$, multiply it by $\hat{\rho }(\epsilon z)$ with a unit-mass mollifier. The product becomes rapidly decreasing in real frequency, so its inverse Fourier transform has an absolutely convergent contour integral. Shifting in a separating direction forces that smooth inverse to vanish outside $K+\epsilon {\overline{B}}_1$. Letting $\epsilon$ tend to zero recovers a distribution supported in $K$.

This Hilbert space has norm $\Vert u{\Vert }_X^2=\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2+\Vert xu{\Vert }_2^2$. Completeness follows from completeness of H1 space and closedness of the multiplication operator $u\mapsto xu$. Smooth cutoffs followed by convolution with a mollifier show that smooth compactly supported functions are dense. The space is the natural energy domain of the dimensionless quantum harmonic oscillator, whereas the operator domain additionally requires $u^{\prime \prime }$ and $x^{2}u$ to lie in L2 space.

Use the Fourier transform convention $\hat{h}(\xi )=\int h(x)e^{-2\pi ix\cdot \xi }dx$. The deleted-ball kernel need not be integrable at infinity, so first work with finite annuli. WriteWe will prove a uniform bound, convergence as $R\to \infty$, and then convergence as $\epsilon \downarrow 0$. This makes all Fourier-transform assertions precise for the annularly cancelling singular convolution kernel.

Fix $\xi \ne 0$ and put $y=\xi /(2|\xi {|}^2)$, so $e^{-2\pi iy\cdot \xi }=-1$. For $R>r\ge 4|y|$, let $\Omega =\{r<|x|<R\}$ and defineThe Hörmander integral kernel condition givesOn changing variables $u=x-y$, the phase factor shows $J_{r,R}=-{\int }_{\Omega -y}K(u)e^{-2\pi iu\cdot \xi }du$. The symmetric difference between $\Omega$ and $\Omega -y$ lies in the two shells whose radii are within $|y|$ of $r$ and $R$. On each such shell, the kernel's size bound gives, by polar integration,Thereforeand adding the two comparisons proves the oscillatory tail estimate for a singular kernel:The integral on the right is at most $B$. Moreover, for fixed $\xi$, it tends to zero as $r\to \infty$ because the translated difference is absolutely integrable outside $2|y|$. The shell term also tends to zero. Thus the far Fourier integral is Cauchy and converges as its outer radius tends to infinity.

Set $r_0=4|y|=2/|\xi |$. If $\epsilon <r_0$, annular cancellation allows subtraction of the constant phase on the near annulus:The same estimate applies to a shorter near annulus when $R<r_0$. The far annulus is bounded by $B/2+C_{d}A$, using the tail estimate; if $\epsilon \ge r_0$, apply that estimate with $r=\epsilon$. At $\xi =0$ the annular integral vanishes by cancellation. We have provedFor each fixed $\epsilon >0$, the size bound makes $K_{\epsilon }\in L^2$, and $K{1}_{\{\epsilon <|x|<R\}}\to K_{\epsilon }$ in $L^2$ as $R\to \infty$. By the Plancherel theorem, their Fourier transforms converge in $L^2$ to ${\hat{K}}_{\epsilon }$. The pointwise improper limits just established identify this transform, up to a null set, withFor example identification follows by selecting an almost-everywhere convergent subsequence from the $L^2$ convergence. HenceThe near-annulus integral with the constant phase subtracted converges absolutely at zero, since its bound is proportional to $|x{|}^{1-d}$. The far integral has already converged. Therefore $m_{\epsilon }(\xi )\to m(\xi )$ for every $\xi \ne 0$, with the same uniform bound; the value at zero may be set arbitrarily.

For $f\in L^2$, the convolution $K_{\epsilon }\ast f$ is absolutely convergent for each $x$ by Cauchy-Schwarz inequality. Its Fourier transform is $m_{\epsilon }\hat{f}$: first prove this for Schwartz functions and then approximate any $L^2$ input. The convolutions converge uniformly under this approximation by the same Cauchy-Schwarz bound, and the multiplier outputs converge in $L^2$ by their uniform operator bound, so these two definitions agree. Define the bounded Fourier multiplier operator $T$ by $\widehat{Tf}=m\hat{f}$. Dominated convergence theorem and the Plancherel theorem now give

For the final request, use a normalized bump function, $\int \varphi =1$, so that its dilates form an approximate identity. Multiplying Fourier transforms givesThese products are integrable by Cauchy-Schwarz inequality, since $T{\varphi }_{\epsilon },f\in L^2$, and also define $L^2$ functions because $m$ and $\hat{\varphi }$ are bounded. Thus the convolution identity is justified even though $T{\varphi }_{\epsilon }$ need not be in $L^1$:The allowed standard approximate identity results, applied to $Tf\in L^2$, prove convergence to $Tf$ in norm and almost everywhere. This is the mass normalization in mollified singular integrals; it does not require proving almost-everywhere convergence of the sharp truncations $T_{\epsilon }f$.

If “bump function” means merely smooth with compact support, the stated limit needs the unit-integral qualification. For arbitrary such $\varphi$, the limit is $(\int \varphi )Tf$ instead. This is also the almost-everywhere limit: at each Lebesgue point of $Tf$, subtract $(\int \varphi )Tf(x)$ inside the convolution and bound the remaining local average by the boundedness and compact support of $\varphi$. As a concrete counterexample without normalization, take $K(x)=1/(\pi x)$ in dimension one and $\varphi =2\rho$ for a unit-integral mollifier $\rho$. This kernel has the required size, odd annular cancellation, and translation-difference integral $\log (3)/\pi$ on $\{|x|>2|y|\}$ for every $y\ne 0$. Its limit operator is the nonzero Hilbert transform; by the Hilbert-transform Fourier multiplier, it is an $L^2$ isometry. For any nonzero $f\in L^2$ the mollified limit is $2Tf$, not $Tf$.

Let $Mf$ be the uncentered maximal function of $|f|$, defined by its averages over balls. First establish the strong $L^p$ estimate needed below. The uncentered maximal weak-type inequality gives the weak $L^1$ bound with $C_d=3^d$, and plainly $Mf\le \Vert f{\Vert }_{\infty }$ for bounded inputs. For $f\in L^p$, split it into its parts on $\{|f|>a/2\}$ and $\{|f|\le a/2\}$. The first part is in $L^1$, since $|f|1_{\{|f|>a/2\}}\le (a/2{)}^{1-p}|f{|}^p$, and the second part contributes at most $a/2$ to the maximal function. Sublinearity therefore givesIntegrating this bound with layer cake representation and Tonelli theorem yieldsThis proves the strong Lp bound from weak L1 and L-infinity bounds directly. In particular $Mf$ is finite almost everywhere.

Now split the positive Riesz potential of $|f|$ at a radius $r$. In the near region, decompose into annuli $2^{-j-1}r<|x-y|\le 2^{-j}r$. On the $j$th annulus,The geometric series converges because $\alpha >0$. In the far region, Hölder's inequality, with $p^{\prime }=p/(p-1)$, givesThe radial integral is finite exactly because $p<d/\alpha$. Combining these bounds, including the fixed Riesz normalization constant, givesFor $0<Mf(x)<\infty$, choose $r=(\Vert f{\Vert }_p/Mf(x){)}^{p/d}$. If $Mf(x)=0$, every ball average vanishes and $f=0$ almost everywhere, so the assertion is trivial. The zero-norm case is likewise trivial. The resulting Hedberg inequality isThis proves absolute convergence of the defining integral for almost every $x$, also for signed or complex $f$. Put $\theta =\alpha p/d\in (0,1)$. The proposed exponent satisfies $q=p/(1-\theta )$, so $(1-\theta )q=p$. Raising the pointwise estimate to the $q$th power and using the proved maximal bound givesThus the Hardy–Littlewood–Sobolev inequality follows from the estimates just derived, rather than being quoted in place of its proof.

For uniqueness of the target exponent, suppose such a uniform estimate holds with index $r$ instead. Choose a nonzero nonnegative bump function $f$ and let $f_t(x)=f(tx)$ for $t>0$. Homogeneity of the Riesz potential givesThe assumed estimate at $t=1$ makes the nonzero output's $L^r$ norm finite. For all $t>0$ it must therefore satisfywith $d/r=0$ when $r=\infty$. Letting $t$ tend to zero or infinity forces the power of $t$ to vanish:Hence no other target index is possible.

For the Sobolev embedding, take $d>1$ and $1<p<d$, so the preceding result is available with $\alpha =1$. For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the vector field $(x-y)/|x-y{|}^d$ has zero divergence in $y$ away from $x$. Integrating by parts outside the ball $|x-y|\le \epsilon$, its inner boundary contributes $|{\mathbb{S}}^{d-1}|$ times the spherical average of $u$ about $x$. The outer boundary contributes zero because $u$ has compact support. Letting $\epsilon \downarrow 0$ provesThe absolute value is at most $C{I}_1(|\nabla u|)(x)$. ThusFor a general member of the first-order Sobolev space $W^{1,p}({\mathbb{R}}^d)=L_1^p({\mathbb{R}}^d)$, approximate in the Sobolev norm by compactly supported smooth functions. This density of smooth functions in a Sobolev space follows from cutoffs tending to one and subsequent mollification: the cutoff-gradient term tends to zero in $L^p$, and a mollifier approximates both the function and each weak derivative. The displayed estimate makes the approximants Cauchy in $L^{q_1}$. Their $L^p$ limit is the original function, so subsequences converging almost everywhere identify the $L^{q_1}$ limit with it. This proves continuous embedding. The range $1<p<d$ is essential for this consequence; the formula is not a general critical $p=d$ embedding into $L^{\infty }$.

The statement as printed is false without an integrability condition. On any nonempty open set, the constant functions $f_n\equiv n$ are harmonic and therefore subharmonic, but their supremum is $+\infty$, which is not locally integrable.

A valid version assumes $g={\sup }_n{f}_n\in L_{\mathrm{loc}}^1(\Omega )$. First, the maximum of finitely many subharmonic functions is subharmonic. For a distributional proof, mollify two functions on smaller interior domains and replace their maximum byThis is smooth, convex, and nondecreasing in each argument. The chain rule shows that its Laplacian applied to two smooth subharmonic functions is nonnegative: the first-derivative terms multiply their nonnegative Laplacians, and the Hessian term is nonnegative by convexity. Mollifier convergence in $L_{\mathrm{loc}}^1$ and the uniform bound $|M_{\delta }-\max |\le \delta /2$ allow both regularizations to be removed. Continuity of distributional derivatives under this convergence proves the finite-maximum assertion.

Now $g_N={\max }_{n\le N}{f}_n$ increases to $g$ and satisfies $|g_N|\le |f_1|+|g|$ almost everywhere. The dominated convergence theorem gives convergence in $L_{\mathrm{loc}}^1$. For every nonnegative test function,Thus $g$ is subharmonic in the distributional sense. This is the locally integrable supremum theorem for subharmonic functions. For the pointwise upper semicontinuous convention, take its canonical representative; the raw supremum need not already have that regularity.

This is Weyl lemma, with the representative condition needed for pointwise smoothness. Choose a smooth radial nonnegative mollifier ${\rho }_{\epsilon }$ of mass one. On interior domains, $f_{\epsilon }=f\ast {\rho }_{\epsilon }$ is smooth and harmonic because $\Delta f_{\epsilon }=(\Delta f)\ast {\rho }_{\epsilon }=0$.

Fix a smaller interior neighborhood and a radial mollifier ${\rho }_r$ whose support stays inside the domain there. The mean value property for harmonic functions implies $f_{\epsilon }=f_{\epsilon }\ast {\rho }_r$ when $\epsilon$ is sufficiently small: a radial average is a weighted average of spherical means, all equal to the center value.

As $\epsilon \downarrow 0$, $f_{\epsilon }\to f$ in $L_{\mathrm{loc}}^1$, while $f_{\epsilon }\ast {\rho }_r\to f\ast {\rho }_r$ locally uniformly. The latter follows by bounding the difference by $\Vert {\rho }_r{\Vert }_{\infty }$ times the local $L^1$ error on a slightly larger compact set. Thus $f=f\ast {\rho }_r$ almost everywhere locally. The right side is smooth, and its Laplacian is zero, providing a smooth harmonic representative. These representatives agree on overlapping neighborhoods since continuous functions agreeing almost everywhere agree everywhere.

The given uniqueness of the canonical upper semicontinuous representative identifies this smooth function with $\tilde{f}$. Since the question assumes $f=\tilde{f}$, it follows that $\boxed{f\in C^{\infty }(\Omega )}$ pointwise. An arbitrary null-set modification of a harmonic distribution would not have that conclusion.

As usual, the bounded set is assumed Lebesgue measurable and an $L^q$ exponent is positive; the conventional Banach-space assertion concerns $1\le q<2^{\ast }=2n/(n-2)$. No boundary regularity of $\Omega$ is needed. Choose a smooth compactly supported cutoff $\eta$ equal to one on a ball containing $\Omega$. Weak convergence and the Uniform boundedness principle give a common $H^1$ bound for $f_j$, hence also for $g_j=\eta f_j$. These functions have a common compact support.

We prove their local $L^2$ compactness directly. Translation and the fundamental theorem along line segments give, by density from smooth functions,For a compactly supported mollifier ${\rho }_{\epsilon }$, integration of this estimate givesAt fixed $\epsilon$, Hölder's inequality bounds $\Vert g_j\ast {\rho }_{\epsilon }{\Vert }_{\infty }$ and $\Vert \nabla (g_j\ast {\rho }_{\epsilon }){\Vert }_{\infty }$ uniformly by the $L^2$ norm of the corresponding smoothing kernels. Their supports lie in one fixed compact set, so the Arzelà-Ascoli theorem makes this smoothed family precompact in $L^2$. The uniform $O(\epsilon )$ approximation makes the original family totally bounded in $L^2$: choose a finite net for the smoothed family and add the approximation error. Completeness gives relative compactness.

Every strongly convergent subsequence of $g_j$ has limit $\eta f$, since multiplication by $\eta$ preserves weak $L^2$ convergence. Consequently the whole sequence converges strongly: otherwise a subsequence with distance at least a fixed positive number would have a further convergent subsequence and contradict that unique limit. In particular $\Vert {\chi }_{\Omega }(f_j-f){\Vert }_2\to 0$.

The critical Sobolev inequality supplies a uniform bound in $L^{2^{\ast }}$. For clarity, this estimate follows from part 2(a), not from compactness at the critical exponent. For a compactly supported smooth $v$, integrating by parts against the vector field $(x-y)/|x-y{|}^n$ givesIts distributional divergence is the sphere-area multiple of a point mass; applying the Hardy–Littlewood–Sobolev inequality with $\alpha =1,p=2$ yields $\Vert v{\Vert }_{2^{\ast }}\le C\Vert \nabla v{\Vert }_2$. Density extends this to $H^1({\mathbb{R}}^n)$.

For $2<q<2^{\ast }$ choose $0<\theta <1$ with $1/q=\theta /2+(1-\theta )/2^{\ast }$. Hölder's inequality givesFor $1\le q<2$, use $\Vert v{\Vert }_{L^q(\Omega )}\le |\Omega {|}^{1/q-1/2}\Vert v{\Vert }_{L^2(\Omega )}$. The same integral estimate also works for $0<q<1$ if $L^q$ is understood with its usual quasi-norm. Thus ${\chi }_{\Omega }{f}_j\to {\chi }_{\Omega }f$ strongly for every positive $q<2^{\ast }$. We have never asserted that multiplying by the possibly nonsmooth indicator preserves $H^1$.

For a real locally integrable function, the definitions are distributional:Thus the Laplacian is respectively a positive distribution, a negative distribution, or zero. A superharmonic function is the negative of a subharmonic function; being both is equivalent to being harmonic. For a $C^2$ function these conditions are $\Delta f\ge 0$, $\Delta f\le 0$, and $\Delta f=0$ pointwise.

An $L_{\mathrm{loc}}^1$ equivalence class does not itself specify point values. For the maximum principle use the canonical representative of a subharmonic distributionThese normalized integrals are ball averages. To justify this construction, first mollify $f$: its smooth local regularizations satisfy $\Delta f_{\epsilon }\ge 0$. For any smooth subharmonic function, the spherical mean $m_x(r)$ obeysThis follows by differentiating the spherical integral and applying the divergence theorem. Its ball mean also increases with radius: its derivative is $n/r$ times the spherical mean minus the ball mean. Local $L^1$ convergence of the regularizations passes the monotonicity of the ball means to $f$. The decreasing-radius limit therefore exists, possibly as minus infinity at an exceptional point, and equals $f$ almost everywhere by the Lebesgue differentiation theorem. It is upper semicontinuous as the local infimum of continuous ball averages and satisfiesFor a superharmonic function use the corresponding lower semicontinuous representative and reverse the inequality.

For a distributionally harmonic function, the smooth regularizations are harmonic and have constant spherical means. A radial mollifier ${\rho }_{\delta }$ of integral one therefore reproduces them by averaging these means. At fixed interior radius $\delta$, pass to the local $L^1$ limit to obtain $f=f\ast {\rho }_{\delta }$ almost everywhere on a smaller region. The right side is smooth and has zero distributional Laplacian, giving the smooth harmonic representative. This proves the relevant Weyl lemma. Changing a value at a single point can spoil a pointwise maximum principle, so the representative convention is essential.

Fix a compact $K\subset \Omega$ and choose $\delta >0$ such that a closed $3\delta$-neighbourhood of $K$ is compactly contained in $\Omega$. The hypothesis gives a common bound $M$ on that neighbourhood. Choose a smooth radial mollifier ${\rho }_{\delta }$ supported in $B_{\delta }(0)$ with integral one. The mean value property for harmonic functions, integrated against its radial weights, givesin a neighbourhood of $K$. Differentiating this fixed-radius convolution gives the uniform interior estimateThus the sequence is equicontinuous there; taking its pointwise limit gives the same local modulus of continuity for $u$. A finite small-ball cover of $K$ now proves uniform convergence directly: choose the cover so both $u_j$ and $u$ vary by less than $\epsilon$ inside each ball, then use convergence at its finitely many centres to make $|u_j-u|<\epsilon$ there. The triangle inequality gives ${\sup }_K|u_j-u|<3\epsilon$ for large $j$.

Finally, pass to the limit in the reproducing convolution identity on smaller compact sets. It yields $u=u\ast {\rho }_{\delta }$, so $u$ is smooth locally. Also, for each compactly supported smooth test function $\varphi$, uniform convergence on its support givesHence its smooth Laplacian is zero. We have proved $u_j\to u$ uniformly on every compact subset and $u$ is harmonic. Using the continuous representatives in the pointwise hypothesis makes this conclusion pointwise, rather than only almost everywhere. This is compact convergence of locally bounded harmonic functions.

The interior elliptic regularity assertion for the Laplace equation is that a harmonic function is smooth, in fact real analytic, throughout $U$. No boundary regularity of its unspecified boundary values is implied.

First let $u\in C^2(U)$ and $\Delta u=0$. On a ball compactly contained in $U$, differentiating its spherical average and applying the divergence theorem expresses that derivative as a constant factor times ${\int }_{B_r}\Delta u$, which is zero. The spherical average tends to $u$ at the center as $r\to 0$. This proves the mean value property for harmonic functions.

Choose a radially symmetric smooth mollifier ${\eta }_r$, supported in $B_r$, with integral one. By integrating the spherical mean value property,whenever $B_r(x)⋐U$. For fixed $r$ the right side is a smooth convolution, since all derivatives can be placed on ${\eta }_r$. Thus $u$ is smooth. The same argument proves the Weyl lemma for a distributionally harmonic $u\in L_{\mathrm{loc}}^1$: first mollify $u$, apply the fixed-radius identity, and let the mollification radius tend to zero in distributions to obtain the same smooth representative.

To prove real analytic regularity, differentiating the fixed-radius convolution gives the interior derivative estimate for a harmonic functionfor any harmonic $v$. All derivatives of $u$ are harmonic. On nested balls between $B_R(a)$ and $B_{R/2}(a)$, apply this estimate $k$ times, decreasing the radius by $R/(2k)$ each time. For $|\alpha |=k$,The inequality $k^k\le e^{k}k!$ follows by integrating $\log s$ below the sum defining $\log (k!)$. Apply the one-dimensional Taylor theorem along each segment, expanding directional derivatives by the multinomial formula. The remainder is bounded by $M(A{\sum }_j|h_j|{)}^k$, so it tends to zero for sufficiently small $h$. This gives a locally convergent multivariate Taylor series. A harmonic function is real analytic in the interior.

The usual inhomogeneous elliptic regularity statement also follows: if $\Delta u=f$ and $f$ is smooth, take a cutoff $\chi$ equal to one near a given point and set $w=\Phi \ast (\chi f)$, where $\Phi$ is a fundamental solution of the Laplace equation with $\Delta \Phi =\delta$. Moving every derivative to the compactly supported smooth function $\chi f$ shows $w$ is smooth. Locally $u-w$ is harmonic, so $u$ is smooth there too.

The construction is well defined. Since $D$ is contained in a ball, its Brownian exit time is finite almost surely: at successive integer times there is a fixed positive probability that the next independent unit-time Brownian motion increment has length greater than the ball's diameter, forcing an exit. The survival probability is therefore bounded by a geometric sequence. Path continuity gives $B_{{\tau }_D}\in \partial D$. Thus the bounded Dirichlet boundary data give a bounded Borel function $u$ on $D$.

For interior harmonicity, fix a ball with closure in $D$, centred at $x$, and let $\sigma$ be its Brownian exit time. The Strong Markov property, followed by the tower property of conditional expectation, givesThe orthogonal invariance of Brownian motion makes $B_{\sigma }$ uniform on the boundary sphere. Hence $u$ has the spherical mean value property for harmonic functions for every such ball. A locally bounded Borel function with this property is smooth and harmonic: integrating the spherical averages against any smooth radial mollifier gives $u=u\ast \rho$ locally, which first proves smoothness; the mean value property for harmonic functions then implies $\Delta u=0$.

For the boundary limit at $\xi$, choose the harmonic barrier for the Dirichlet problem $h_{\xi }$ from (i). The Itô formula, localized inside $D$, makes $h_{\xi }(B_{t\wedge {\tau }_D})$ a bounded martingale. Compact localization and the bounded convergence theorem, first to the exit and then as $t\to \infty$, yieldFor $\epsilon >0$, if $\partial D\setminus B(\xi ,\epsilon )$ is nonempty, compactness and barrier positivity giveIf that boundary subset is empty, the probability is already zero. ThereforeFirst let $x\to \xi$ and then $\epsilon \downarrow 0$. The continuity of the Dirichlet boundary data proves $u(x)\to f(\xi )$.

Finally, the difference of two continuous solutions is a harmonic function vanishing on the boundary. The maximum principle for harmonic functions on the bounded domain gives that it is zero. Thus the Kakutani solution of the Dirichlet problem exists, attains every prescribed boundary value, and is unique.

Reflect $u$ across the two endpoints of $I$ into a slightly larger interval, giving an $H^1$ extension, and multiply by a smooth cutoff function equal to one near $\overline{I}$. Convolve the resulting function on $\mathbb{R}$ with a mollifier. The convolutions $u_{\epsilon }$ are smooth; differentiation commutes with convolution of the weak derivative. Continuity of translations in $L^2$ proves both $u_{\epsilon }\to u$ and $u_{\epsilon }^{\prime }\to u^{\prime }$ in $L^2(I)$, hence convergence in the Sobolev norm.

To obtain convergence almost everywhere, select a sequence with $\Vert u_n-u{\Vert }_2\le 2^{-n}$. Then $\int {\sum }_n|u_n-u{|}^2<\infty$, so the summands tend to zero almost everywhere. Thus a smooth approximating sequence converges both in $H^1$ and almost everywhere. This is density of smooth functions in a Sobolev space; smooth functions need not vanish at the endpoints.

The Weak Harnack inequality is an estimate for a nonnegative weak supersolution. Write $F=(f^1,\dots ,f^n)$, let $B_{2R}(x_0)\subset \subset \Omega$, and assume $u\ge 0$ almost everywhere on this ball. There are $p>0$ and $C<\infty$, depending only on the dimension, the uniform ellipticity bounds, $q$, and the scaled norms of the lower-order coefficients, such thatHere $⟨v{⟩}_E=|E{|}^{-1}{\int }_{E}v$ is the integral average. The weak supersolution inequality uses the sign convention $Lu\le \mathrm{div}F+g$. For instance, the coefficient dependence can be expressed usingThus one may use the same $p,C$ on all sufficiently small balls when the global coefficient norms and uniform ellipticity bounds are fixed. The nonnegativity condition is essential to this formulation; a signed weak supersolution may first be shifted, with the resulting change in its forcing included. As usual, supremum and infimum statements for functions in a Sobolev space mean essential suprema and essential infima. The standard multidimensional statement uses $n\ge 2$, so $q/2>1$. In dimension one the analogous formulation requires $q\ge 2$; the printed restriction $q>n$ alone would allow $q/2<1$, which does not suffice. To see the obstruction, smooth the nonnegative capped function $\min \{1,|x|/\epsilon \}$ by a mollifier of width $\tau$. Its positive second derivative is a bump of mass $2/\epsilon$ and has $L^{q/2}$ size of order ${\epsilon }^{-1}{\tau }^{2/q-1}$. For fixed $\epsilon$ and $1<q<2$, this size and the minimum both tend to zero as $\tau \downarrow 0$, whereas the average on a fixed larger interval stays positive. Thus the Weak Harnack inequality cannot have the displayed uniform forcing bound in that range. Shrinking $\epsilon$ and choosing $\tau$ still smaller likewise defeats a uniform Hölder estimate based on that norm alone.

On a bounded Lipschitz domain $\Omega \subset {\mathbb{R}}^d$, the Rellich-Kondrachov compactness theorem states that $W^{1,p}(\Omega )\hookrightarrow L^q(\Omega )$ is compact for $1\le p<\infty$ andwith $1/q=0$ for $q=\infty$. Thus $q<p^{\ast }$ when $p<d$, every finite $q$ is allowed when $p=d$, and the embedding into continuous functions with their uniform norm is compact when $p>d$.

For the main proof, a Sobolev extension operator puts a bounded sequence into a common compactly supported region of ${\mathbb{R}}^d$. The Sobolev fundamental theorem of calculus on lines gives the uniform translation estimate $\Vert u(\cdot +h)-u{\Vert }_p\le |h|\Vert \nabla u{\Vert }_p$. Convolution with a mollifier therefore approximates the sequence uniformly in Lp space. At any fixed smoothing scale, its derivatives and supremum are uniformly bounded, so the Arzelà-Ascoli theorem gives a convergent subsequence. A diagonal choice and the uniform approximation give strong Lp space convergence. The finite measure of $\Omega$ handles $q<p$, while Lp interpolation inequality with the bounded Sobolev embeddings upgrades this to the larger subcritical finite Lp spaces; for $p>d$, Morrey's inequality gives uniform equicontinuity directly and the Arzelà-Ascoli theorem gives uniform convergence. Boundedness of the domain is essential.