# Is the parameter-dependent integral of a Sobolev function continuous?

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.

• 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"? Jul 17, 2021 at 15:16
• @Nate Yes, exactly! I've edited the question Jul 17, 2021 at 15:47
• 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. Jul 17, 2021 at 16:09

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 $$F \in W^{1,p}_{\mathrm{loc}}(\mathbf{R})$$.
• You're correct, that's what I was referring to here: first (1) for $f$ then (2) for $F$. Jul 18, 2021 at 16:11