# $\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ implies $u\in L^{6/5}(\Omega)$

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.

• You are asking if $H^{-1} \subset L^{6/5}$, which is false – Piero D'Ancona Jan 8 '16 at 0:24
• @Piero D'Ancona Can you give me some argument or a reference, so that I can see why is this? – Svetoslav Jan 11 '16 at 14:11
• 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 $x_1$). – Piero D'Ancona Jan 11 '16 at 14:23
• @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 $v(x)\in H_0^1$ – Svetoslav Jan 11 '16 at 15:15

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 $v\in H_0^1$.
• 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 ? – Svetoslav Jan 9 '16 at 19:18
• 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 – Svetoslav Jan 10 '16 at 20:58
• 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. – Svetoslav Jan 10 '16 at 21:06
• 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. – Christian Remling Jan 12 '16 at 21:46
• @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 $\int |uv|=\int\liminf |uv_n|\le\liminf\int|uv_n|\lesssim \liminf\|v_n\|=\|v\|$. – Christian Remling Jan 13 '16 at 18:20