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I posted this question first in Math.StackExchange one week ago here, but I didn't get an answer or a helpful comment so I repost it here:

Let $d=3$ and $\Omega\subset \mathbb R^d$ is a bounded Lipschitz domain and $u$ is a measurable function. A sufficient condition for the integral $\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ is that $u\in L^{6/5}(\Omega)$ which follows from Holder's inequality and the (continuous) embedding $H^1(\Omega)\hookrightarrow L^6(\Omega)$.

Question: Is the opposite true, i.e is it true that $$\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$$ implies $u\in L^{6/5}(\Omega)$ or at least $u\in L^1(\Omega)$ ?

My thoughts: It is easy to see that $u\in L^1_{loc}(\Omega)$ by taking $v$ to be smooth cut-off functions equal to $1$ in compact subsets of $\Omega$ and $0$ in a neighborhood of the boundary $\partial \Omega$.

The motivation for this question is the "correct" weak formulation of a nonlinear problem - whether to formulate it as $(1)$ or as $(2)$:

$(1)$ Find $u\in H_0^1(\Omega)$ such that $f(u)\in L^{6/5}(\Omega)$ and $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$

or

$(2)$ Find $u\in H_0^1(\Omega)$ such that $\int\limits_{\Omega}{f(u)vdx}<\infty,\forall v\in H_0^1(\Omega)$ and $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$

where $a(.,.)$ is a bilinear form and $f(.)$ is in general a nonlinear function. If the answer to my question is affirmative then both formulations are equivalent.

Note that $(2)$ is less restrictive, because the set in which we search for a solution is bigger, so it might be easier to find such.

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    $\begingroup$ You are asking if $H^{-1} \subset L^{6/5}$, which is false $\endgroup$
    – Piero D'Ancona
    Commented Jan 8, 2016 at 0:24
  • $\begingroup$ @Piero D'Ancona Can you give me some argument or a reference, so that I can see why is this? $\endgroup$
    – Svetoslav
    Commented Jan 11, 2016 at 14:11
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    $\begingroup$ Take $u=\partial_1(\chi(x)v(x_1))$ where $\chi$ is a test function on $R^3$ equal to 1 near the origin and $v(s)$ is the sign function. From $\chi v\in L^2$ it follows $u\in H^{-1}$, and of course $u$ is not even a function since it is a distribution (a delta in the direction $x_1$). $\endgroup$
    – Piero D'Ancona
    Commented Jan 11, 2016 at 14:23
  • $\begingroup$ @Piero D'Ancona Yes, this is fine, but my functional is not just any functional (distribution) but it is a regular distribution, i.e it is represented by an integral of the measurable function $u(x)$ times a function $v(x)\in H_0^1$ $\endgroup$
    – Svetoslav
    Commented Jan 11, 2016 at 15:15

1 Answer 1

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That doesn't work because $H_0^1$ functions are small near the boundary, so testing against them won't detect bad behavior of $u$ near $\partial\Omega$.

For a concrete example, take $\Omega$ as the unit ball and $u(x)=1/(1-|x|)\notin L^1$. Then $$ \int |uv|\, dx \le \left( \int \frac{v^2\, dx}{(1-|x|)^{3/2}} \int \frac{dx}{(1-|x|)^{1/2}} \right)^{1/2} . $$ If $v\in H_0^1$ is also smooth, then we can estimate the first integral in the same way as in this related question (by just integrating the gradient, starting from the boundary, to bound $v$). This gives $\int v^2/(1-|x|)^{3/2}\lesssim \|v\|^2_{H^1}$, so $\int |uv| \lesssim \|v\|_{H^1}$ for all such $v$, and by density of the smooth functions, this also holds for arbitrary $v\in H_0^1$.

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  • $\begingroup$ sorry for the late response. I like your answer, but I cann't get the crucial inequality $|v(x)|\leq (1-|x|)^{1/2}\|v\|_{H_0^1}$. Could you elaborate a bit on it ? $\endgroup$
    – Svetoslav
    Commented Jan 9, 2016 at 19:18
  • $\begingroup$ If I understand correctly the issue, I think that this inequality can not be true (for $C_0^\infty$ functions), because (I think) there are positive functions $f\in C_0^\infty(B_0(1))$ with arbitrary small $H^1$ norm and arbitrary "high": $\forall M>0\forall \delta>0\exists f\in H_0^1(B_0(1)): \|f\|_{H^1(B_0(1))}<\delta$ and $\exists \epsilon>0: f(x)\ge M$ a.e in $B_0(\epsilon)\subset B_0(1)$. For example, if we take $g=K \log{\frac{1}{|x|/\epsilon}}$ and construct the function $\tilde g=g$ on $B_0(\epsilon)$ and $\tilde g=0$ on $B_0(1)\setminus B_0(\epsilon)$. Here $K$ controls the $H^1$-norm $\endgroup$
    – Svetoslav
    Commented Jan 10, 2016 at 20:58
  • $\begingroup$ Then, if $h(x)=\tilde g(x)$ where $x: \tilde g(x)\leq M$ and $h(x)=M$ where $x: \tilde g(x)\ge M$. Now taking $f$ to be a mollified version of $h$ should give us $C_0^\infty(B_0(1))$ function, for which the inequality does not hold. Maybe I am missing something, but it will be nice if you check my construction. $\endgroup$
    – Svetoslav
    Commented Jan 10, 2016 at 21:06
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    $\begingroup$ Notice that the $t$ integration is essentially the same thing as integrating with respect to the radial variable in spherical coordinates, so integrating over just the angular part of $dx$ gives you a quantity that is estimated by $\|v\|^2$, and then the extra $d|x|$ integration just affects the constant. $\endgroup$
    – Christian Remling
    Commented Jan 12, 2016 at 21:46
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    $\begingroup$ @Svetoslav: I think Fatou's Lemma works best here: we also have that $v_n\to v$ pointwise a.e. (pass to a subsequence if necessary), so $\int |uv|=\int\liminf |uv_n|\le\liminf\int|uv_n|\lesssim \liminf\|v_n\|=\|v\|$. $\endgroup$
    – Christian Remling
    Commented Jan 13, 2016 at 18:20

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