Suppose $f:\Bbb R^2\to \Bbb R$ is twice continuously differentiable, and that $f, f_{xx}, f_{yy}\in L^2$. Then is it true that $f$ is bounded?
I first tried to find a counterexample, but I couldn't. So it seems true (maybe) but how can we show this? Thanks in advance.
On
This is true. Since $f_{xx}$ and $f_{yy}$ belong to $L^2$, the Laplacian $\Delta f$ is in $L^2$. By elliptic regularity, $f\in H^2=W^{2,2}$. By Sobolev inequality you have $$ \frac 1{p}-\frac k{n}=\frac 1{2}-\frac 2{3}<0 $$ and therefore $f\in L^\infty$. I do not know how familiar you are with PDEs though.
This is an Answer that merely expands what GReyes said. Some basic functional analysis, the definition of Sobolev spaces, and a little Fourier analysis is all that is needed. (If you knew the $L^p$ results and Sobolev theory as laid out in say Evans, I think you would agree that this is elementary.)
Choose the Fourier transform convention $\widehat f(\xi) = \mathcal Ff (\xi) = \int f(x) e^{-2\pi i x\xi} dx$ which is a Schwartz function, if $f$ is. The Fourier transform is extended to $f\in L^2$ by the density of Schwartz functions and Plancharel’s theorem $\|\mathcal F f\|_{L^2} = \|f\|_{L^2}$ to a unitary map $\mathcal F : L^2 \to L^2$. The inverse is $\mathcal F^{-1} g(x) = \mathcal Fg(-x)$.
Elliptic Regularity
The $k$th Riesz transform $R_k$ is the Fourier multiplier defined by the $L^\infty(\mathbb R^2)$ function $m_k(\xi) = -i \xi_k/|\xi|$, $$\widehat{ R_k f }= m_k \widehat f.$$ Since $\frac12=\frac1\infty + \frac12$, and $\|m_k\|_{L^\infty} = 1$, Holder’s inequality(the easiest case, where you just pull the supremum out of the integral) immediately proves that $\widehat{ R_k f}$ is an $L^2$ function, and then applying the inverse transform, $R_k $ is bounded $L^2 \to L^2$. For Schwartz functions $f$, it’s easy to see that $f_{xy}=f_{yx}=R_1R_2\Delta f$. So at least for such $f$, we have $$\|f_{xy}\|_{L^2} \le \| \Delta f\|_{L^2}. $$ Also, integration by parts and Cauchy- Schwarz gives (again for Schwartz functions $f$)
$$ \|\nabla f\|^2_{L^2} = \int_{\mathbb R^2} \nabla f\cdot \nabla f = -\int_{\mathbb R^2} f\Delta f \le \|f\|_{L^2} \|\Delta f\|_{L^2}$$
Now take a general $f$ as in the question and a sequence of Schwartz functions $f^n$ such that $\Delta f^n \to \Delta f$ and $f^n\to f$ in $L^2$. ($n$ is an index, not a power.) Then by the first inequality, $f^n_{xy}$ is Cauchy and therefore convergent in $L^2$ to some $g$. It is not too hard to check that $g=f_{xy}$ in the sense of weak derivatives: $$ \int_{\mathbb R^2} f \phi_{xy} = \lim_{n\to\infty} \int_{\mathbb R^2} f^n \phi_{xy} = \lim_{n\to\infty}\int_{\mathbb R^2} f^n_{xy} \phi =\int_{\mathbb R^2} g\phi.$$ Similarly $\nabla f^n$ converges to some vector field $h$ in $L^2$, which can be shown to be the weak gradient of $f$. So this proves that $f\in H^2$ (i.e. $f$ has all weak derivatives up to second order, and they are all $L^2$ functions).
Sobolev Embedding (Morrey's inequality)
Now we prove a small part of a special case of Morrey's inequality, which proves that $H^2(\mathbb R^2)$ continuously embeds into the Banach space of $1/6$-Hölder continuous functions. We will just prove boundedness, since that is all that you wanted. Staring at Plancharel for a few moments reveals that $\int(1+|\xi|^2)^2 |\mathcal F f(\xi)|^2 d\xi \le C^2 \|f\|_{H^2}^2$ for some $C>0$. For any Schwartz $f$ and any $x$, we have $$ f(x) = \mathcal F^{-1} \mathcal F f = \int_{\mathbb R^2} \mathcal F f(\xi) e^{2\pi i x \xi} d\xi = \int_{\mathbb R^2}\frac1{1+|\xi|^2}(1+|\xi|^2) \mathcal F f(\xi) e^{2\pi i x \xi} d\xi $$ So by Cauchy-Schwarz, \begin{align} |f(x)| &\le \int_{\mathbb R^2}\frac1{1+|\xi|^2}(1+|\xi|^2) |\mathcal F f(\xi)| d\xi \\&\le \underbrace{\sqrt{\int \frac{d\xi}{(1+|\xi|^2)^2}}}_{=C'<\infty} \sqrt{ \int(1+|\xi|^2)^2 |\mathcal F f(\xi)|^2 d\xi} \\&\le CC' \|f\|_{H^2}. \end{align} Since $x$ was arbitrary, this proves that $\|f\|_{L^\infty}\le CC' \|f\|_{H^2}$, for Schwartz functions. By the approximation argument again, we see that a general $f$ as determined in the question is in fact bounded on $\mathbb R^2$, as was requested. (In fact, the continuity assumption on $f$ can be removed, by the previously mentioned Morrey's inequality.)