# $u_{xx}$ and $u_{yy}$ continuous $\Rightarrow u$ is continuous

I found an exciting question in a scriptum:

Let $U\subset\mathbb{R}^2$ open and convex, $u:U\to\mathbb{R}$. Assume the existence of the continuous partial derivatives $u_{yy}$ and $u_{xx}$ satisfying $u_{yy}=u_{xx}$. Note that the existence of the mixed partial derivatives is not explicitly assumed.

Can I conclude that the function $u$ is continuous? Otherwise should I need more assumptions?

More explicitly, let $f(x)=u(x,0)$, $g(x)=u_y(x,0)$. Then $f\in C^2(\mathbb R)$, $g\in C^1(\mathbb R)$ by assumption. We have that $$2u(x,y)= f(x-y)-G(x-y) + f(x+y)+G(x+y) ,$$ with $G'=g$. To confirm this, we can fix $x$ and then check that $u$ as a function of $y$ has the correct second derivative and the right initial values $u(x,0)=f(x)$, $u_y(x,0)=g(x)$.
We conclude that in fact $u\in C^2(\mathbb R^2)$; in particular, $u$ is continuous.
Fix $(x_0,y_0)\in U$. For small enough $\varepsilon$ we have $u(x_0+\varepsilon,y_0)+const = \int_{x_0}^{x_0+\varepsilon}\int_{x_0}^{t} u_{xx}(s,y_0)\ ds\,dt.$ Since $|u_{xx}|$ is bounded on small enough retangle around $(x_0,y_0)$ the Lebesgue Dominated convergence applies and so when $\varepsilon\to 0$ we have that $\lim_{\varepsilon\to 0}u(x_0+\varepsilon,y_0)=u(x_0,y_0)$. Similarly you can prove the continuity on the second variable, so $u$ has to be continuous on both variables.
• Continuity with respect to both variables is not enough to conclude continuity. E.g. $f(x,y) = xy/(x^2+y^2)$ for $(x,y)\ne(0,0)$, and $f(0,0) = 0$. Jan 19, 2016 at 0:27
• @Richard my argument probably is not fixable because it is not using $u_{xx}=u_{yy}$ which I guess is crucial. I am keeping in hope someone can fix it. Jan 19, 2016 at 17:35