Let $\Omega = \mathbb{R}^3$ (could consider $[-L,L]^3$ for simplicity if needed) and consider the space $H^1(\Omega)$ of functions with square integrable functions and gradients. Let $\|\cdot\|_{H^1(\Omega)}$ be defined as $$\|v\|^2_{H^1(\Omega)} = \int_{\Omega} \|\nabla v(x)\|^2 dx + \int_{\Omega} v(x)^2 dx$$ where $\|\cdot\|$ is the euclidean norm. I am trying to show that the bilinear form $a(u,v)$ defined as $$a(u,v) = \int_{\Omega} \nabla u(x) \cdot \nabla v(x) dx - \int_{\Omega} \frac{1}{\|x\|} u(x) v(x) dx$$ satisfies the following inequality for some positive constants $\gamma_1$ and $\gamma_2$: $$\|v\|^2_{H^1(\Omega)} \leq \gamma_1 a(v,v) + \gamma_2 \|v\|_{L^2(\Omega)}^2.$$ For now, I have that \begin{align*} \|v\|^2_{H^1(\Omega)} & \leq a(v,v) + \int_{\Omega} \frac{1}{\|x\|}v(x)^2dx + \|v\|_{L^2(\Omega)}^2 \\ & \leq a(v,v) + \|\nabla v\|_{L^2(\Omega)}\|v\|_{L^2(\Omega)} + \|v\|^2_{L^2(\Omega)} \end{align*} where I used the Coloumb uncertainty principle: $$\int_{\Omega} \frac{1}{\|x\|}v(x)^2dx \leq \|\nabla v\|_{L^2(\Omega)}\|v\|_{L^2(\Omega)}.$$ I am wondering if there are any upper bounds on $\|\nabla v\|_{L^2}$ in terms of $\|v\|_{L^2}$ that I could use? That would complete the proof, but perhaps there may be another way to reach my desired conclusion.


In one dimension take $\Omega = [-1,1]$ and let $v_k = \dfrac{\sin (k^2 x)}{k}$. In this case you unfortunately get $\|v_k\|_{L^2(\Omega)} \to 0$ and $\|\nabla v_k\|_{L^2(\Omega)} \to \infty$. This is easily adaptable to higher dimensions.

$$ \|v\|_{H^1(\Omega)}^2 - 2 a(v,v) = \|v\|_{L^2(\Omega)}^2 - \|\nabla v\|_{L^2(\Omega)}^2 + 2 \int_\Omega \frac 1{\|x\|} v(x)^2 \, dx $$ and $$ \int_\Omega \frac 1{\|x\|} v(x)^2 \, dx \le \|\nabla v\|_{L^2(\Omega)} \|v\|_{L^2(\Omega)} \le \frac 1{2} \|\nabla v\|_{L^2(\Omega)}^2 + \frac 1{2} \|v\|_{L^2(\Omega)}^2 $$ so that $$ \|v\|_{H^1(\Omega)}^2 - 2 a(v,v) \le 2 \|v\|_{L^2(\Omega)}^2.$$

  • $\begingroup$ Ah that's a good example. My approach seems infeasible then. Any ideas on how I can bound the $\int 1/\|x\| v^2$ term? I know of Hardy's inequality but that gives a bound for $\int 1/\|x\|^2 v^2$ which isn't exactly what I have. $\endgroup$ – xk3 Apr 13 '18 at 19:06
  • $\begingroup$ Your approach simply needs an adjustment of the constants. See the edit. $\endgroup$ – Umberto P. Apr 16 '18 at 15:54

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