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An approximate identity is a family of integrable kernels $K_{\epsilon }$ whose total mass is one and whose mass concentrates near zero as $\epsilon \to 0$. Under standard hypotheses, $f\ast K_{\epsilon }$ converges to $f$ in norm and at almost every Lebesgue point.

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.

If a locally integrable function has weak derivatives represented by a continuous vector field $g$, it has a representative $U\in C^1$ with $\nabla U=g$. The assertion is local, so work on a ball compactly contained in the domain and regularize using a mollifier. The regularizations satisfy the segment identity $u_{\epsilon }(x)-u_{\epsilon }(y)={\int }_0^1\nabla u_{\epsilon }(y+t(x-y))\cdot (x-y)dt$. Averaging over a smaller ball in $y$ fixes the additive constant through local $L^1$ convergence. The gradients converge uniformly on compact sets, so the averaged identity gives local uniform convergence to $U$. Passing to the limit on each segment then gives $U(x+h)-U(x)-g(x)\cdot h=o(|h|)$. Continuity of $g$ proves that $U$ is continuously differentiable.

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.

A distribution $u$ on a plane open set with ${\partial }_{\bar{z}}u=0$ is represented by a holomorphic function. Indeed $\Delta u=4{\partial }_z{\partial }_{\bar{z}}u=0$, and elliptic regularity makes $u$ smooth. For a direct proof, mollify locally with a radial mollifier ${\eta }_{\epsilon }$. The resulting smooth functions are harmonic. Their mean value property gives $(u\ast {\eta }_{\epsilon })\ast {\eta }_r=u\ast {\eta }_{\epsilon }$ on a smaller ball for fixed small $r$. Passing to the weak convergence of distributions limit gives $u\ast {\eta }_r=u$ there. This proves smoothness without assuming local integrability beforehand. The ordinary Cauchy-Riemann equations now prove holomorphy.

The equation ${\partial }_{z}u=0$ similarly characterizes antiholomorphic functions. On the whole plane, the homogeneous solutions are respectively arbitrary entire functions $h(z)$ and the functions $h(\bar{z})$ with $h$ entire. Consequently any two fundamental solutions for one of these Wirtinger derivatives differ by the corresponding homogeneous solution.

Every $C^2$ harmonic function is locally the real part of a holomorphic function and is therefore infinitely differentiable.

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.

The weak convergence of distributions $u_j\to u$ meansFor $\theta \in \mathcal{D}(\mathbb{R})$, its convolution with a distribution is the smooth functionThe subscript records the variable on which the distribution acts. With $\check{\theta }(x)=\theta (-x)$, its regular distribution satisfies the useful pairing identityIndeed integrating $\theta (x-y)\varphi (x)$ in $x$ gives $(\check{\theta }\ast \varphi )(y)$; continuity of the distribution justifies moving the pairing through that integral, by approximation with sums of test functions in one common compact support.

Part (i), applied to the reflected mollifier, gives ${\check{\rho }}_{\epsilon }\ast \varphi \to \varphi$ in the space of test functions. Continuity of $u$ therefore provesReflection is harmless even if the chosen standard mollifier is not even.

For the mollifiers themselves, a change of variable givesThus $\boxed{{\rho }_{\epsilon }\to {\delta }_0\text{ in }{\mathcal{D}}^{\prime }(\mathbb{R})}$, with limit the Dirac delta distribution, despite the failure of convergence in the L1 space.

Convergence ${\varphi }_j\to \varphi$ in the space of test functions means that all supports lie in one compact set $K\subset \mathbb{R}$ and $\Vert {\varphi }_j^{(r)}-{\varphi }^{(r)}{\Vert }_{\infty }\to 0$ for every integer $r\ge 0$. An eventual common compact support is equivalent, since finitely many initial supports can be added to $K$.

Write the standard mollifier as ${\rho }_{\epsilon }(x)={\epsilon }^{-1}\rho (x/\epsilon )$, where $\rho \in \mathcal{D}(\mathbb{R})$ is nonnegative, supported in $[-1,1]$, and has integral one. For $0<\epsilon \le 1$, the convolution ${\rho }_{\epsilon }\ast \varphi$ has support in the fixed compact set $\mathrm{supp}\varphi +[-1,1]$. The given convolution derivative identity yieldsEach ${\varphi }^{(r)}$ is uniformly continuous; more explicitly, the mean value theorem bounds the absolute value by $\epsilon \Vert {\varphi }^{(r+1)}{\Vert }_{\infty }\int |y|\rho (y)dy$, uniformly in $x$. Thus

If ${\varphi }_j\to 0$ in the space of test functions and $K$ is a common compact support, then $\Vert {\varphi }_j{\Vert }_{L^1}\le |K|\Vert {\varphi }_j{\Vert }_{\infty }\to 0$, where $|K|$ is its Lebesgue measure.

The mollifiers themselves have no limit in the L1 space. If ${\rho }_{\epsilon }\to g$ in that norm, then for each $\delta >0$ the integral of $|g|$ on $\{|x|>\delta \}$ is zero, because ${\rho }_{\epsilon }$ vanishes there for $\epsilon <\delta$. Hence $g=0$ almost everywhere. This contradicts $\Vert {\rho }_{\epsilon }{\Vert }_{L^1}=1$. Therefore $\boxed{{\rho }_{\epsilon }\text{ does not converge in }{L}^1(\mathbb{R})}$.

If $F$ is any other fundamental solution, then $H=F-E$ satisfies ${\partial }_{\bar{z}}H=0$ as a distribution. Applying $4{\partial }_z$ gives $\Delta H=0$. Elliptic regularity for the Laplacian implies that the initially arbitrary distribution $H$ is a smooth function. Its equation is then the ordinary Cauchy-Riemann equation, so $H$ is an entire function.

For a direct justification of the needed regularity, choose a radial mollifier ${\eta }_r$. Locally, $H\ast {\eta }_{\epsilon }$ is a smooth harmonic function. On a ball with sufficient room for both radii, the mean value property gives $(H\ast {\eta }_{\epsilon })\ast {\eta }_r=H\ast {\eta }_{\epsilon }$. Passing to the weak convergence of distributions limit $\epsilon \to 0$ yields $H\ast {\eta }_r=H$ on a smaller ball. The left side is smooth, proving the claim locally. This establishes the distributional Cauchy-Riemann lemma without assuming in advance that $H$ is a locally integrable function.

Conversely every entire function $h$ satisfies ${\partial }_{\bar{z}}h=0$ and can be added to $E$. Hence all fundamental solutions areNo growth condition at infinity was imposed, so $h$ need not be a polynomial.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }({\mathbb{R}}^3)$ with $\int \rho =1$, and put ${\rho }_{\epsilon }(x)={\epsilon }^{-3}\rho (x/\epsilon )$. Sobolev approximation by mollifiers in the Sobolev space under consideration statesIn particular, it gives density of smooth functions in a Sobolev space. We prove convergence of both the function and all its first weak derivatives.

The usual smoothing properties of convolution give $u_{\epsilon }\in C^{\infty }$. For a test function $\varphi \in C_c^{\infty }({\mathbb{R}}^3)$ and $i\in \{1,2,3\}$, Fubini's theorem and the definition of a weak derivative giveThe integrals are absolutely convergent: the compact supports of ${\rho }_{\epsilon }$ and $\varphi$ restrict $y$ to a compact set, on which $u$ and ${\partial }_{i}u$ are integrable by Holder inequality. ConsequentlyThe approximate identity theorem in the Lp space $L^6$ applies separately to $u$ and ${\partial }_{i}u$. ThusEach convolution belongs to $L^6$ by Young's convolution inequality. With the componentwise Sobolev norm, the sixth power of the required difference is the sum of these four sixth powers, proving the boxed claim. Any equivalent Sobolev norm gives the same convergence.

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.