Show existence of a continuous $k$ on $\mathbb{R}^2$ such that $(u,\phi)= \int_{\mathbb{R}^2}k\phi dx$ for all $\phi$

Question

(b). Let $u$ be a distribution on $\mathbb{R}^2$. Assume there exists a continuous function $h$ on $\mathbb{R}^2$ such that $(u,\Delta \phi) = \int_{\mathbb{R}^2}h\phi dx $ for all $\phi\in C_0^{\infty}(\mathbb{R}^2)$. ( $\Delta = \partial_x^2+\partial_y^2$)

Show there exists a continuous function $k$ on $\mathbb{R}^2$ such that $(u,\phi)= \int_{\mathbb{R}^2}k\phi dx$ for all $\phi$.

In (a) Ive shown $H_{\text{loc}}^2(\mathbb{R}^2) \subset C^0(\mathbb{R}^2)$ which will probably be of relevant importance. ($H_{\text{loc}}^2$ is the set of distributions $u\in D'$ such that $\phi u\in H^2$ for all $\phi$).

So how to show (b)? Tips or suggestions are greatly appreciated. Thanks.

Answer 1

Let $f\in D'(\mathbb{R}^2)$ be the fundamental solution of the Laplace equation (in fact: $f(x) = C \ln ||x||$), so $\Delta f = \delta$. Let $h_n$ be a function equal to $h$ on $B_n$, the ball of radius $n$, and $0$ outside the $B_{n+1}$. Then $f_n := f\ast h_n$ is a continuous function and $\Delta f_n = h_n$.

So $u-f_n$ is a harmonic distribution on $B_n$, hence by the well-known result of H. Weyl in fact a harmonic $C^\infty$-function. Restricted to $B_n$, as $f_n\in C^0(B_n)$, this implies $u\in C^0(B_n)$.

As continuity is a pure local property, u equals a continuous function everywhere.

Answer 2

Here's an answer for a related, but simpler question. Instead of $\mathbb{R}^2$ I look at $\mathbb{R}$, and instead of $\Delta = \partial_x^2 + \partial_y^2$, I take the differentiation operator $\frac{d}{dx}$.

You then have $(u,\phi) = (u,\Delta\Phi) = \int h\Phi$ where $\Phi(x) = \int_0^x \phi(t) \,dt$. Changing the order of integration yields \begin{align} (u,\phi) &= \int_{-\infty}^\infty h(x)\Phi(x) dx = \int_{-\infty}^\infty h(x) \int_{-\infty}^x \phi(t) \,dt\,dx = \int_{-\infty}^\infty \int_t^\infty h(x)\phi(t)\,dx\,dt \\ &= \int (1-H)\phi \quad\text{where}\quad H(x) = \int_{-\infty}^x h(t)\,dt \text{.} \end{align}

It should be possible to do basically the same thing for your differential operator, and for $\mathbb{R}^2$ instead of $\mathbb{R}$, I think.