I am trying to prove that the following distribution is not given by a continuous function.
Having $f$ a continuous function on $\mathbb{R}$. For $\varphi\in C_{c}^{\infty} (\mathbb{R}^{2})$, we would define $<u,\varphi>=\int_{\mathbb{R}}\varphi (x,f(x))dx $. I was able to prove that $u$ is a distribution in R squared, with order 0 and support equals to the point of the type $(x,f(x))$.
Then I was asked to prove that $u$ is not given by a continuous function, i.e. that for all $\tilde{u}\in C(\mathbb{R}^{2})$ there exists $\varphi\in C_{c}^{\infty} (\mathbb{R}^{2})$ such that $<u,\varphi>\neq \int \tilde{u}(x,y)\varphi (x,y)dxdy$ and I stuck here.
Would like some help how to prove that. Thanks
Let $$G = \{(x,\,f(x))\in\mathbb R^2: x \in \mathbb R\}$$ denote the graph of $f$. Because $f$ is continuous, $G$ is a closed set. Now suppose we did have some $\tilde u$ continuous. It clearly cannot be the zero function. Without loss of generality, assume that $$P = \{ (x,y) \in \mathbb{R}^2\mid \tilde u(x,y) > 0\} \neq \emptyset$$ (otherwise, just look at the set $\{\tilde u<0\}$). The set $P$ is open by the continuity of $\tilde u$. Then we have that $P\setminus G$ is also an open set.
With an appropriate choice of $\varphi\in C_c^\infty(\mathbb R^2)$, we can reach a contradiction...
We can pick any $\mathbf x \in P\setminus G$ and choose $\epsilon$ sufficiently small such that $B_{\epsilon}(\mathbf x)$ (the ball of radius $\epsilon$ centered at $\mathbf x$) is contained in $P\setminus G$. We can construct $\varphi\in C^\infty_c(\mathbb R^2)$ where:
On the one hand $$\langle u, \varphi\rangle = \int_{\mathbb R} \varphi(x,f(x)) \, dx = 0,$$ since $\varphi \equiv 0$ outside of $B_{\epsilon}(\mathbf x) \subset P\setminus G$, i.e., $\varphi$ is exactly zero on the graph of $G$. On the other hand, $$ \int_{\mathbb R^2}\tilde u\,\varphi \, dx\, dy \geq \int_{\overline{B_{\epsilon/2}(\mathbf x)}} \tilde u \, dx\, dy > 0,$$ since $\tilde u > 0$ on $\overline{B_{\epsilon/2}(\mathbf x)}\subset P\setminus G$. Therefore $$\langle u, \varphi\rangle \neq \int_{\mathbb R^2}\tilde u\,\varphi \, dx\, dy.$$