Let $w_m=|E_m{|}^{5/6}{u}_m$. By linearity of the weak formulation, $w_m$ is the zero-boundary weak solution with forcing ${\chi }_{E_m}$. The estimate in (a)(iii) therefore givesThe change of variables $x=r_{m}y$ gives $|E_m|=r_m^3|E|>0$, so the division is legitimate. Since $H^1(B)$ is a Hilbert space, weak sequential compactness in a Hilbert space now gives a subsequence and some $u\in H^1(B)$ such thatIn fact $u\in H_0^1(B)$: the Sobolev trace operator is continuous and linear, so its zero kernel is preserved by weak convergence.

By the definition in (i), $u$ itself is an admissible test in the weak formulation. HenceThe absolute value makes a proof of positivity unnecessary. Holder inequality with exponents $6$ and $6/5$, followed by the zero-boundary Sobolev inequality, givesTaking square roots of both estimates proves exactly the requested chain:If $\Vert \nabla u{\Vert }_2=0$, the zero-boundary Poincaré inequality gives $u=0$ and the conclusion follows. Otherwise divide the first-to-last estimate by $\Vert \nabla u{\Vert }_2^{1/2}$ and square, obtaining $\Vert \nabla u{\Vert }_2\le C_S|E{|}^{5/6}$. Finally the same Poincaré inequality givesAll constants depend only on the fixed unit ball and the dimension, and are independent of $E$ and $u$. If $|E|=0$, testing the homogeneous equation with $u$ gives the same zero solution directly.

As printed, the answer isThe constant function $u\equiv 1$ has zero gradient and positive Lp norm for every such $q$, including its supremum norm when $q=\infty$.

The missing assumption is a zero Sobolev trace, which is present for the weak solutions in the surrounding parts. With $H_0^1(B)$ in place of $H^1(B)$, the intended answer isThis is the three-dimensional Sobolev inequality at $q=6$, followed by Holder inequality on the bounded ball for smaller $q$; it includes the zero-boundary Poincaré inequality at $q=2$. The subsequent parts use this zero-boundary version.

The zero Dirichlet boundary condition must be part of the definition. Thus $u\in H^1(B)$ is a weak solution precisely when $u\in H_0^1(B)$, equivalently its Sobolev trace vanishes on $\partial B$, andHere ${\chi }_E$ is the indicator function of the measurable set $E$. One may initially require the identity only for test functions $\varphi \in C_c^{\infty }(B)$: both sides are continuous in the Sobolev norm, and the definition of the zero-trace Sobolev space then extends it to all $H_0^1(B)$. This formulation uses first weak derivatives and encodes $-\Delta u={\chi }_E$ without assuming classical second derivatives.

Set $w=(u-k{)}^{+}$, the positive part of $u-k$. The allowed Sobolev chain rule givesUse the admissible zero-boundary Sobolev space function ${\phi }^{2}w$ in the weak formulation, with a valid cutoff function as explained in (ii). Its gradient is ${\phi }^2\nabla w+2\phi w\nabla \phi$, soIndeed the integrands vanish off $\{u>k\}$, and on that set $\nabla w=\nabla u$. The uniform ellipticity and Cauchy-Schwarz inequality estimates from (ii) now apply to $w$ and giveFixing the cutoff function gives the required constant independent of $k$. This is a superlevel Caccioppoli inequality; it uses the positive part as a test and does not require the truncated function itself to solve the original equation.

First repeat the Caccioppoli inequality calculation on two concentric balls. If $h$ is a harmonic function on $B_R$, choose a cutoff function equal to one on $B_r$, supported in $B_R$, and with gradient bounded by $C/(R-r)$. The argument of (ii), with $A=I$ and $k=0$, givesHere $C_{\ast }$ depends at most on the dimension. When $h$ is only known to be smooth inside $B_R$ and in $L^2(B_R)$, the weak formulation and all integrals involving the cutoff function are still legitimate, since its support is strictly inside $B_R$.

Every classical partial derivative of the given smooth harmonic function $u$ is again a harmonic function, since constant-coefficient partial derivatives commute with the Laplacian. Choose indices $i_1,\dots ,i_m$ representing the multi-index $\alpha$, and defineFor each $j\ge 1$, the preceding Caccioppoli inequality yieldsStarting with $u\in L^2(B_1)$, these estimates also establish the integrability needed at each subsequent step. Iterating $m$ times givesThis is the desired interior derivative estimate for a harmonic function, with a constant depending only on $n$ and $m$, uniformly over every multi-index of size $m$.

There is a small incompatibility in the printed cutoff function assumptions. No $\phi \in C_c^{\infty }(B_1)$ can equal one on $B_{1/2}$ and satisfy $|\nabla \phi |\le 2$: for a unit vector $e$, choose $t<1$ beyond its support. Continuity gives $\phi (e/2)=1$, while the fundamental theorem of calculus would give $1\le 2(t-1/2)<1$. Replace the number $2$ by a fixed sufficiently large absolute constant $C_0$; such a cutoff function exists, and the requested Caccioppoli inequality is unchanged apart from its constant. We keep $M=\Vert \nabla \phi {\Vert }_{\infty }$ in the calculation, which also yields the formal bound under the printed assumptions.

The weak formulation is ${\int }_{B_1}A\nabla u\cdot \nabla \zeta =0$ for test functions $\zeta$. Since $A$ is bounded, this extends by density of smooth functions in a Sobolev space to every $\zeta \in H_0^1(B_1)$. Part (a)(ii), followed by mollification inside the compact support, makes ${\phi }^2(u-k)$ an admissible choice. The product rule givesPut $X=\int {\phi }^2|\nabla u{|}^2$ and $Y=\int (u-k{)}^2|\nabla \phi {|}^2$. The lower and upper uniform ellipticity bounds, together with Cauchy-Schwarz inequality, implyIf $X=0$ the conclusion is immediate. Otherwise divide by $X^{1/2}$ and square to obtain $X\le 4{\Lambda }^2{\lambda }^{-2}Y$. Since $\phi =1$ on $B_{1/2}$,For a fixed valid cutoff function, $M\le C_0$, this has the required constant independent of $k$. Formally inserting the printed value $M=2$ gives $16{\Lambda }^2/{\lambda }^2$. No derivative of $A$ was taken, so differentiability of the coefficients is unnecessary.

Interpret the paper's unspecified $\Omega$ as containing $B_1$, the domain on which the coefficients are used. Write $A(x)=(a_{ij}(x))$, a real symmetric matrix. The strict ellipticity condition is pointwise positive definiteness:The lower bound may depend on $x$. By continuity, strict ellipticity gives a positive lower bound on each compact subset of $B_1$, but does not require one bound throughout the whole open ball.

With the two-constant convention used in the next part, uniform ellipticity meansBoth constants are independent of $x$. Some terminology uses uniform ellipticity for just the lower bound and states coefficient boundedness separately; here the paper supplies both constants. The spectral theorem for the symmetric matrix gives $|A(x)\xi |\le \Lambda |\xi |$. For example, $A(x)=(1-|x{|}^2)I$ satisfies strict ellipticity in $B_1$ but has no uniform positive lower bound as $|x|\uparrow 1$.

For $\psi \in C_c^{\infty }(B_1)$, the functions $\psi$ and ${\partial }_i\psi$ are bounded. Hence $\psi u$ and $\psi {\partial }_{i}u+u{\partial }_i\psi$ belong to $L^p(B_1)$. For a test function $\varphi$, the classical product rule and the weak derivative identity for $u$, tested with $\psi \varphi$, giveThereforeIn fact $\psi u$ has compact support in $B_1$ and belongs to the corresponding zero-boundary Sobolev space, by mollification within that support.

For arbitrary $\psi \in C^{\infty }(B_1)$, the same identity still proves the local product rule for weak derivatives, but membership in the global Sobolev space can fail. To give a counterexample for every dimension and every $p$ in the question, takeHere $u\in W^{1,p}(B_1)$ and $\psi$ is a smooth function on the open ball, while polar coordinates showThus $\psi u\notin L^p(B_1)$, and the global assertion is false. Boundedness of $\psi$ and its first derivatives would suffice without compact support.

A locally integrable function $u$ has weak derivative $g_i$ in direction $i$ when $g_i$ is locally integrable andThe test functions have compact support inside the domain, so no boundary values enter this definition. Such $g_i$ is unique almost everywhere and is denoted ${\partial }_{i}u$.

For $1\le p<\infty$, the first-order Sobolev space consists of the Lp space classes whose first weak derivatives are also in that Lp space:This Sobolev norm is equivalent to $\Vert u{\Vert }_p+\Vert \nabla u{\Vert }_p$; either convention is acceptable. When $p=2$, this is the Hilbert space $H^1(B_1)$.

Use the continuous representative of $\bar{u}$ furnished by the one-dimensional Sobolev embedding theorem, and define $u^{\ast }(x)=\bar{u}(|x|)$ on $A$. The one-dimensional Sobolev representative is absolutely continuous; Holder inequality applied to its fundamental theorem of calculus givesAlso $\Vert \bar{u}{\Vert }_{\infty }\le \Vert \bar{u}{\Vert }_6+\Vert {\bar{u}}^{\prime }{\Vert }_6$: select a point where $|\bar{u}|$ is no larger than its $L^6$ norm on the unit-length interval and apply the same fundamental theorem of calculus. Thus the one-dimensional Hölder norm is controlled by $\Vert \bar{u}{\Vert }_{W^{1,6}(1,2)}$.

The two Hölder seminorms must be compared using the correct distances. By the reverse triangle inequality,Conversely, take $x=re$ and $y=se$ for any fixed unit vector $e$. Then $|x-y|=|r-s|$, so taking suprema yields equality of the two Hölder seminorms. Their supremum norms are equal as well. Part (i) now givesThe representative agrees with $u$ almost everywhere, and the assignment is linear and injective on the radial Sobolev space. This proves the requested continuous embedding. The inner radius being positive makes the radial profile Sobolev norm comparable to its weighted radial version.

Write $r=|x|$, $u(x)=\bar{u}(r)$, and let $g(r)$ be the radial profile of the radial derivative $\nabla u(x)\cdot x/|x|$, which may be assumed to be a radial function here. The spherical coordinate system givesSince $1\le r^2\le 4$, both $\bar{u}$ and $g$ belong to the Lp space $L^6(1,2)$. It remains to identify the weak derivative of the radial profile directly.

For $\varphi \in C_c^{\infty }(1,2)$, each component of $\varphi (r)x/r$ is a test function on $A$. Sum the three defining weak derivative identities, using the stated divergence formula:Given an arbitrary test function $\zeta \in C_c^{\infty }(1,2)$, choose $\varphi (r)=\zeta (r)/r^2$. Dividing by $4\pi$ gives $\int g\zeta =-\int \bar{u}{\zeta }^{\prime }$, exactly the definition of the weak derivative. We have provedThis radial Sobolev profile on an annulus estimate follows from the two weighted integrals above and equivalence of the usual Sobolev norms.

Here Morrey's inequality has Hölder exponent $1-3/6=1/2$. Every $u\in W^{1,6}({\mathbb{R}}^3)$ has a unique continuous representative $u^{\ast }$ such thatWe use the Hölder norm $\Vert f{\Vert }_{C^{0,1/2}}=\Vert f{\Vert }_{\infty }+[f{]}_{C^{0,1/2}}$, where the Hölder seminorm is ${\sup }_{x\ne y}|f(x)-f(y)|/|x-y{|}^{1/2}$. The second bound is the homogeneous form of Morrey's inequality; the first includes control of the values themselves.

Let $u_j={\rho }_{1/j}\ast u$ be the mollifications from (i). They converge to $u$ in the Sobolev norm. Apply the assumed smooth version of Morrey's inequality to the differences $u_j-u_{\ell }$:Thus $(u_j)$ is a Cauchy sequence in the complete Hölder space $C^{0,1/2}({\mathbb{R}}^3)$ and converges there to a function $w$. In particular the convergence is uniform. On every bounded set $K$ of finite volume,Exhausting ${\mathbb{R}}^3$ by balls proves $w=u$ almost everywhere. Set $u^{\ast }=w$. Taking limits in the smooth Hölder norm bound gives the first boxed estimate. For any fixed $x\ne y$, taking limits ingives the second boxed estimate after taking the supremum. Finally, two continuous representatives equal almost everywhere must coincide everywhere, since a nonzero continuous difference would remain nonzero on an open ball of positive volume.

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 intended space of Hilbert-Schmidt operators, with the given Hilbert-Schmidt inner product linear in its first argument. It remains to prove completeness in its Hilbert-Schmidt norm. Let $(T_n)$ be a Cauchy sequence in that norm. The bound $\Vert T_n-T_m\Vert \le \Vert T_n-T_m{\Vert }_2$ makes it a Cauchy sequence in operator norm. The space $\mathcal{B}(X)$ is a Banach space, so $T_n\to T$ in operator norm for some bounded linear operator $T$.

For completeness, that last standard fact follows by taking $Tx={\lim }_n{T}_{n}x$ for each $x$: the Cauchy sequence $(T_n)$ is uniformly bounded in operator norm, so its limit is linear and bounded, and the uniform Cauchy estimate on the unit ball gives operator norm convergence.

Given $\epsilon >0$, choose $N$ so that $\Vert T_n-T_m{\Vert }_2<\epsilon$ for $n,m\ge N$. Fix $n\ge N$ and any positive integer $K$. Since $T_m{e}_k\to T{e}_k$ for each $k$, passage to the limit in a finite sum givesTaking the supremum over $K$ gives ${\sum }_k\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$. Also,Thus $T\in S_2(X)$, and $\Vert T_n-T{\Vert }_2\le \epsilon$ for $n\ge N$. This proves Hilbert-Schmidt operators form a Hilbert space:Under the literal unrestricted “linear mappings” definition, the nonzero $R$ from part (i) satisfies $⟨R,R{⟩}_{S_2(X)}=0$. The form would fail positive definiteness and would not be an inner product; the boundedness correction is needed here too.

As in the preceding part, take $T$ to be a bounded linear operator. For $x\in X$, the Hilbertian basis expansion and continuity of $T$ giveThe triangle inequality, the Cauchy-Schwarz inequality for finite sums, and the Parseval identity for a Hilbertian basis givePassing to the limit and taking the supremum over the unit ball provesThus the Hilbert-Schmidt norm controls the operator norm. Without the boundedness interpretation, the nonzero $R$ constructed in the preceding part has zero printed column sum and makes the requested inequality false.

Here the intended space must consist of bounded linear operators, namely Hilbert-Schmidt operators. The printed restriction to “all linear mappings” is insufficient. To see this, choose $v\in X$ outside the algebraic linear span of $(e_k)$, for example $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$. Extend the linearly independent family $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis. Define a linear functional $\varphi$ on this Hamel basis by $\varphi (v)=1$ and zero on every other basis vector, and extend it linearly. Then the everywhere-defined linear operator $Rx=\varphi (x)e_1$ satisfiesIts printed column sum is zero although $R\ne 0$. More explicitly, $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k\to v$ but $R(v-v_N)=e_1$ for every $N$. Thus $R$ is discontinuous; it has no everywhere-defined adjoint operator of the bounded kind used in the question. This proves finite column sum does not imply boundedness. We use $T\in \mathcal{B}(X)$ in the intended argument below.

By the Parseval identity for a Hilbertian basis for $(e_j)$, followed by the adjoint operator identity and then the Parseval identity for a Hilbertian basis for $(h_k)$,The interchange is valid by Tonelli's theorem for nonnegative double series, even before finiteness is known. Repeating with $h_k=e_k$ shows that the right side also equals ${\sum }_k\Vert T{e}_k{\Vert }^2$. HenceIn particular the Hilbert-Schmidt norm is independent of the Hilbertian basis, and $\Vert T^{\ast }{\Vert }_2=\Vert T{\Vert }_2$.

Again put $A=\lambda I-T$. Surjectivity gives, by image-kernel orthogonality for an adjoint,If $\lambda$ is real, $A=A^{\ast }$, so $\ker A=\{0\}$. If $\lambda$ is nonreal and $Ax=0$, then $Tx=\lambda x$; reality of $⟨Tx,x⟩=\lambda \Vert x{\Vert }^2$ again implies $x=0$. Thus $A$ is injective as well as surjective, without needing to assume separately that $\lambda$ is not an eigenvalue.

Since $A$ is a bounded linear operator on the Banach space $X$, the bounded inverse theorem makes $A^{-1}$ bounded. ThereforeThis is exactly membership in the resolvent set of an operator.

Put $A=\lambda I-T$. Since $T$ is a self-adjoint operator, $A^{\ast }=\overline{\lambda }I-T$. Applying the preceding image-kernel orthogonality for an adjoint identities to $A^{\ast }$ givesIf $\lambda$ is real, $A^{\ast }=A$ and $\ker A=\{0\}$ because $\lambda$ is not an eigenvalue. If $\lambda$ is nonreal and $A^{\ast }x=0$, then $Tx=\overline{\lambda }x$. The scalar $⟨Tx,x⟩$ is real by self-adjointness, while it equals $\overline{\lambda }\Vert x{\Vert }^2$. Its imaginary part forces $x=0$. Thus $\ker A^{\ast }=\{0\}$ in both cases, andThe conjugation on $\lambda$ in the adjoint operator is essential when $\lambda$ is nonreal.

Let $J:C([0,1])\to L^1(0,1)$ send a continuous function to its almost everywhere equivalence class. The domain with the supremum norm and the target with the L1 norm are Banach spaces, andThus $J$ is a bounded linear operator, and the requested vector subspace is $J(C([0,1]))$.

To show this image of a linear map is not closed, consider the continuous functionsand $g=1_{(1/2,1)}$. Their L1 norm difference is the area of one triangle:But $g$ has no continuous function representative. Indeed, a continuous function equal to zero almost everywhere on $(0,1/2)$ must be identically zero there: any nonzero value would persist on an open interval of positive Lebesgue measure. Likewise any continuous representative would be identically one on $(1/2,1)$. Continuity at $1/2$ makes these requirements incompatible. Therefore $g\notin J(C([0,1]))$, although it lies in its closure.

The preceding operator range category dichotomy proves continuous functions are meagre in L1: