Let $f\in W^{1,2}\_{\text{loc}}(\mathbb R^2)$. Here, $W^{1,2}\_{\text{loc}}(\mathbb R^2)$ denotes the usual Sobolev space. More explicitly, $f:\mathbb R^2\to\mathbb R$ is a function such that, for every relatively compact open set $U\subset\mathbb R^2$,


* $f\vert\_U\in L^2(U)$ ;
* there exist $g\_1,g\_2\in L^2(U)$ such that $$\int\_U f\partial\_1\phi=-\int\_U g\_1 \phi,\text{ and }\int\_U f\partial\_2\phi=-\int\_U g\_2 \phi$$
for all test functions $\phi\in C\_{\text{c}}^\infty(U)$.




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**My question.** Is the function $F:\mathbb R^2\to\mathbb R$, defined by
$$F(x,y)=\int\_0^y f(x,t)\,\mathrm dt$$
continuous?


More precisely stated, does there exist a function $\tilde F\in C(\mathbb R^2)$ such that $F=\tilde F$ Lebesgue-almost everywhere? (Note that the function $F$ is not well-defined at every point since $f$ is only defined as an equivalence class modulo "being equal almost everywhere".)




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Note that (cf. Brezis *Functional Analysis, Sobolev Spaces and Partial Differential Equations*, Lemma 8.2) the function $F$ is continuous in the variable $y$ and, weakly, $\partial\_2 F=f$. However, I don't even see whether $F$ needs to be continuous in $x$.


*Remark.* If we had for instance $f\in W^{1,3}\_{\text{loc}}(\mathbb R^2)$, then it would be clear that $F$ is continuous, since, by Morrey's inequality (see Evans *Partial Differential Equations*, chapter 5.6.2, Theorem 4), the space $W^{1,p}(\mathbb R^n)$ can be embedded into $C^0(\mathbb R^n)$ whenever $p>n$. But my case is $p=n$, so this Theorem doesn't apply.



Is the parameter-dependent integral of a Sobolev function continuous?
I believe this holds more generally—here is the attempt I propose. Consider a function $f \in W^{1,p}\_{\mathrm{loc}}(\mathbf{R}^2)$ for some $p > 1$. Since the second variable is fixed in the problem, we can take $y = 1$ and define $F(x) = \int\_0^1 f(x,t) \mathrm{d} t$, outside of some negligible subset in $\mathbf{R}$.


The claim is that this function inherits from $f$ the property that
$$
F \in W^{1,p}\_{\mathrm{loc}}(\mathbf{R}),$$
from which the desired conclusion follows.


Let $I \subset \mathbf{R}$ be a finite interval. Then $$\int\_I \lvert F \rvert^p = \int\_I \Big \lvert \int\_0^1 f(x,t) \mathrm{d}t \Big\rvert^p \mathrm{d} x
\leq \int\_I \int\_0^1 \lvert f(x,t) \rvert^p \mathrm{d} t \mathrm{d} x,$$
so that $F \in L\_{\mathrm{loc}}^p(\mathbf{R})$.


In the same vein, given $h \in \mathbf{R}$ let $\tau\_h F: x \mapsto F(x-h)$. Then
\begin{eqnarray\*}
\int\_I \lvert \tau\_h F - F \rvert^p
&=& \int\_I \Big \lvert \int\_0^1 f(x+h,t) - f(x,t) \mathrm{d} t \Big \rvert^p \mathrm{d} x \\
&\leq& \int\_I \int\_0^1 \lvert f(x+h,t) - f(x,t) \rvert^p \mathrm{d} t \mathrm{d} x.
\end{eqnarray\*}
In other words
$$
\lvert \tau\_h F - F \rvert\_{L^p(I)} \leq \lvert \tau\_{he\_1}f - f \rvert\_{L^p(I \times [0,1])},$$
where $\tau\_{he\_1}$ is the translate of $f$ in the direction of the standard basis vector $e\_1 \in \mathbf{R}^2$.


The characterisation of Sobolev functions in terms of difference quotients means that there is $C > 0$ so that $$ \lvert \tau\_{he\_1}f - f \rvert\_{L^p(I \times [0,1])} \leq C h$$ for small enough $h$, which in turn implies that $F \in W^{1,p}\_{\mathrm{loc}}(\mathbf{R})$.


