I am trying to prove the continuity of the functional \begin{equation} f: v \mapsto \int_{\mathbb{R}^N} \vert v \vert^2 \end{equation} on the space \begin{align} V(\mathbb{R}^N) = \lbrace v = v_1 + i v_2 : \mathbb{R}^N \to \mathbb{C} ~ | ~ &\nabla v \in L^2(\mathbb{R}^N), \\ &v_1 \in L^2(\mathbb{R}^N), \\ &v_2 \in L^4(\mathbb{R}^N),\\ & \nabla v_1 \in L^\frac{4}{3}(\mathbb{R}^N) \rbrace, \end{align} equipped with the norm $$ \Vert v \Vert_{V(\mathbb{R}^N)} = \Vert \nabla v \Vert_{L^2(\mathbb{R}^N)} + \Vert v_1 \Vert_{L^2(\mathbb{R}^N)} + \Vert v_2 \Vert_{L^4(\mathbb{R}^N)} + \Vert \nabla v_1 \Vert_{L^\frac{4}{3}(\mathbb{R}^N)}.$$

Now, if I try to prove the continuity I choose $v,w \in V(\mathbb{R}^N)$ such that $$ \Vert v - w \Vert_V < \delta $$ for some $\delta>0$. In particular, $$\Vert \nabla v \Vert_{L^2(\mathbb{R}^N)} = \int \vert \nabla v_1- \nabla w_1 \vert^2+\vert \nabla v_2-\nabla w_2 \vert^2< \delta.$$ I go on to compute \begin{align} \vert f(v) - f(w) \vert &= \left\vert \int \vert \nabla v \vert^2 - \int \vert \nabla w \vert^2 \right\vert \\ &= \left\vert \int \vert\nabla v_1\vert^2 - \vert\nabla w_1\vert^2 + \vert \nabla v_2 \vert^2 - \vert\nabla w_2\vert^2 \right\vert \end{align} Here I'm stuck. How do I show that this is smaller than any $\varepsilon>0$ if only $\delta$ is small enough? Or is there another way to show the continuity? Any hint is much appreciated!

EDIT. In the contrary: $\vert \vert a \vert^2-\vert b \vert^2 \vert \not\leq C \vert a - b \vert^2$ in general (for $N=1$ try $b=a-\varepsilon$).

  • 1
    $\begingroup$ Are you trying to do this in every dimension $N$ or can $N$ be restricted? The reason I'm asking is that since the functional is convex you essentially have to bound the $L^2$ norm by some function of the $V$ norm. The way to do this is usually by some kind of Sobolev inequality, and these tend to depend crucially on the space dimension. $\endgroup$ – user38355 Aug 9 '13 at 14:55
  • $\begingroup$ @brom That's true. Thank you for the hint. Let's assume $N=3$. $\endgroup$ – mjb Aug 9 '13 at 21:17

The $L^2$ norm of $v_1$ is obviously controlled by the $V$-norm. But the $L^2$-norm of $v_2$ is not; that is, the functional is not continuous. Indeed, let $v$ be a nonzero purely imaginary function, $v=iv_2$. For $t>0$ let $v_t=t^{-N/2}v(tx)$. By the change of variables, $$\begin{split}\int_{\mathbb R^N} |v_t|^2 &= \int_{\mathbb R^N} |v|^2 \\ \int_{\mathbb R^N} |\nabla v_t|^2 &= t^{-2} \int_{\mathbb R^N} |\nabla v|^2 \\ \int_{\mathbb R^N} |v_t|^4 &= t^{-N} \int_{\mathbb R^N} |v|^4 \end{split}$$ Therefore, $\|v_t\|_V\to 0$ as $t\to\infty$, while $f(v_t)$ remains the same (nonzero) value.

  • $\begingroup$ Bummer. But thank you very much! $\endgroup$ – mjb Aug 14 '13 at 8:02

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