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A distribution on $(a,b)$ is a continuous linear functional on the space of test functions $\mathcal{D}(a,b)=C_c^{\infty }(a,b)$. It has order of a distribution at most $m$ if, for every compact set $K\subset (a,b)$, there is $C_K$ such thatThe order of a distribution is the least such $m$. When saying “of order $m$” means merely that this estimate holds, the wording allows a smaller least order.

For each such $K$, the cutoff function theorem supplies a nonnegative $\chi \in \mathcal{D}(a,b)$ equal to one near $K$. If $M=\Vert \varphi {\Vert }_{\infty }$ and $\mathrm{supp}\varphi \subseteq K$, then $M\chi \pm \varphi \ge 0$ everywhere. Positivity of the linear functional $V$ therefore givesHere $V(\chi )\ge 0$ is finite and independent of $\varphi$. This is precisely the local bound for an order-zero distribution, so the given positive distribution is continuous on the space of test functions. The zero distribution also satisfies the estimate.

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.

Take the extension from part (a), and choose $M>N+n+1$. The positive functionbelongs to $X_{N,k}$. Positivity extends from nonnegative Schwartz functions to nonnegative elements obtained as $X_{N,k}$ limits. Sincepositivity givesTaking $K=\tilde{v}[w]$ proves that every positive distribution has order zero.