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 $\phi\in C_{\text{c}}^\infty(U)$.

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".)

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.

  • 1
    $\begingroup$ Since $f$ isn't well-defined pointwise in $x$, and neither is $F$, I guess the question is really "does $F$ have a continuous version"? $\endgroup$ Jul 17, 2021 at 15:16
  • $\begingroup$ @Nate Yes, exactly! I've edited the question $\endgroup$ Jul 17, 2021 at 15:47
  • $\begingroup$ This should be true by some version of the trace theorem. Restriction of $f$ to any horizontal line is in $L^2$ continuously depending on that line and so integral is well-defined and continuous as well. $\endgroup$ Jul 17, 2021 at 16:09

1 Answer 1


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 $F \in W^{1,p}_{\mathrm{loc}}(\mathbf{R})$.

  • $\begingroup$ Great! If I understand correctly, in the last paragraph you are using something along the lines of Theorem 1.46 (i) here. $\endgroup$ Jul 18, 2021 at 14:17
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    $\begingroup$ You're correct, that's what I was referring to here: first (1) for $f$ then (2) for $F$. $\endgroup$
    – Leo Moos
    Jul 18, 2021 at 16:11

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