Let $\Omega\subset\mathbb R^3$ be a bounded and Lipschitz domain and let $f:\Omega\to\mathbb R$ be a measurable function such that $$\left|\int\limits_{\Omega}{fvdx}\right|\leq C\|v\|_{H^1(\Omega)},\,\forall v\in H_0^1(\Omega)\cap L^\infty(\Omega)\quad\quad (*)$$ In other words, the measurable function $f$ defines a bounded linear functional $F_f$ on the dense subspace $H_0^1(\Omega)\cap L^\infty(\Omega)\subset H_0^1(\Omega)$ through the formula $$\langle F_f,v\rangle =\int\limits_{\Omega}{fvdx},\,\forall v\in H_0^1(\Omega)\cap L^\infty(\Omega)\quad\quad (**)$$

Question: Is it true that the unique extension $\bar F_f$ (by continuity) of this functional is representable again with the same formula $(**)$ ? So, in my opinion, the question is equivalent to whether the quantity $\int\limits_{\Omega}{fvdx}$ is well defined (finite) for all $v\in H_0^1(\Omega)$ and if finite, whether $$\langle \bar F_f,v\rangle =\int\limits_{\Omega}{fvdx},\,\forall v\in H_0^1(\Omega)$$

My thoughts: It is (clear) that $f\in L_{loc}^1(\Omega)$. This can be seen by choosing $v\in C_0^\infty(\Omega)$ to be smooth cut-off functions equal to $1$ in $K\subset\subset\Omega$ and $0$ in a neighborhood of $\partial \Omega$ and plugging in $(*)$. It is also clear that the extension by continuity is defined by $$\forall v\in H_0^1(\Omega),\\ \langle \bar F_f,v\rangle =\lim\limits_{n\to\infty}{\langle F,v_n\rangle}=\lim\limits_{n\to\infty}{\int\limits_{\Omega}{fv_ndx}},\,\forall \{v_n\}\subset H_0^1(\Omega)\cap L^\infty(\Omega): \|v_n-v\|_{H^1(\Omega)}\to 0$$ where the limit always exists and it does not depend on the approximating sequence. So, the question is reduced to showing that if $v_n\to v$ with $v_n\in H_0^1(\Omega)\cap L^\infty(\Omega)$, then $\int\limits_{\Omega}{fv_ndx}\to\int\limits_{\Omega}{fvdx}$, knowing that the limit exists. It is sufficient to prove the above only for a subsequence of $\{v_n\}$ and thus we can choose $\{v_n\}\subset H_0^1(\Omega)\cap L^\infty(\Omega)$ such that we also have pointwise a.e convergence of $v_n$ and its weak derivatives to that of $v$ and use some convergence theorem.

Question 2:(probably easier) If we have condition $(*)$ but for functions in $C_0^\infty(\Omega)$, does it follow that the extension $\bar F_f$ to the space $H_0^1(\Omega)\cap L^\infty(\Omega)$ is representable with the formula $$\langle \bar F_f,v\rangle =\int\limits_{\Omega}{fvdx},\,\forall v\in H_0^1(\Omega)\cap L^\infty(\Omega)$$ In this case we can choose the approximating sequence $v_n\to v$ such that $v_n(x)\to v(x)$, $\frac{\partial v_n(x)}{\partial x_i}\to v\frac{\partial v(x)}{\partial x_i}$ a.e and $\|v_n\|_{L^\infty(\Omega)}\leq \|v\|_{L^\infty(\Omega)}+1$. However, we still cannot apply Lebesgue DCT, because $|f(x)v_n(x)|\leq |f(x)|\|v_n\|_{L^\infty(\Omega)}$, but $f\notin L^1(\Omega)$.

An answer or a fruitful suggestion on either of the questions would be highly appreciated.

Here is an example for a function $f\notin L^1$ such that $\int\limits_{\Omega}{fvdx}\leq C\|v\|_{H^1(\Omega)},\,\forall v\in H_0^1(\Omega)$.

  • $\begingroup$ If you meant to put an absolute value around the integral of (*), then the answer is yes. Extensions by continuity are unique. So the extension is the obvious one because the obvious one extends and is continuous. $\endgroup$ – DisintegratingByParts Sep 14 '16 at 22:26
  • $\begingroup$ @TrialAndError Thanks for the comment. I think there is no difference whether it has absolute value or not - LHS is linear. Also I cannot understand what you are saying. It is clear that it is uniquely extendable, but my question is: Is the unique extension representable by the same formula $\int\limits_{\Omega}{fvdx}$ ? I.e is this formula still valid for the functions $v\in H_0^1(\Omega)$ which are not in $L^\infty(\Omega)$ ? $\endgroup$ – Svetoslav Sep 14 '16 at 23:11
  • $\begingroup$ I jumped to a conclusion about the function $f$. $\endgroup$ – DisintegratingByParts Sep 15 '16 at 1:59
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    $\begingroup$ @TrialAndError I think I found the answer. If you are interested, look below. $\endgroup$ – Svetoslav Sep 19 '16 at 18:03

I think I found an answer to my question. Essentially the problem above (in different generalizations) is considered in a series of 3 papers by Haim Brezis and Felix Browder. Taking, for example the paper "Sur une propriete des espaces de Sobolev" from 1978, we see that a sufficient condition for that the extension $\bar F_f$ evaluated at an arbitrarily fixed element $v\in H_0^1(\Omega)$ to be representable by the formula $\langle \bar F_f,v \rangle=\int\limits_{\Omega}{fvdx}$ is that $f\in L_{loc}^1(\Omega)$ and that there exists a function $g\in L^1(\Omega)$ such that $f(x)v(x)\ge g(x),\,a.e$ in $\Omega$.

  • $\begingroup$ Very nice. Thanks for calling my attention to this. $\endgroup$ – DisintegratingByParts Sep 19 '16 at 18:35

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