2
$\begingroup$

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.

$\endgroup$
3
  • 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

1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.