I know the density argument in $L^p$ space, in Sobolev spaces, and even in BV are really sweet in many many cases. However, some times author just work on nice functions and comment by "the rest can be easily done by density"... Like I said, some times it is easy, but some times it is really not clear! Here I list some problems I encountered before in which I do not find out why density argument can so easily worked out.

This one regrading to weak convergence in $BV$ space. It is an argument in Evans & Gariepy's book, page 175, theorem 3. In that theorem they were trying to prove a sequence of radon measure $\mu_k\to \mu$ weakly. In the prove, they only test against $\phi\in C_c^1(R^n;R^n)$, and hence they can employ integration by parts in $BV$ functions. However, to my knowledge, to prove a radon measure convergence weakly, we really need to test against all continuous function. So here I assume they use density argument. Now given any $\phi\in C_c(R^n;R^n)$, of course we can find a sequence $\phi_m\in C_c^\infty$ and $\phi_m\to \phi$ uniformly. Hence, the question would be, why we can interchange the limit such that $$\lim_{m\to\infty}\lim_{k\to\infty}\int \phi_m d\mu_k =\lim_{k\to\infty}\lim_{m\to\infty}\int \phi_m d\mu_k \,\,??$$ given that we already proved that $$\lim_{k\to\infty}\int \phi_m d\mu_k =\int \phi_m d\mu $$ for all $\phi_m\in C_c^\infty$.

Another similar question regrading to density is here, in which I have a sequence $u_n$ bounded in $H^1$ norm and I want to prove $u_n\to 0$ weakly in $H^1$. It is quick to show that $u_n\to 0$ weakly in $L^2$ but hard to work out $\partial_i u_n\to 0$. The comment in that post nicely suggest by using density argument so that we could test $\partial_i u_n$ with a $C^\infty$ function and hence by integration by parts we could use that fact $u_n\to 0$ weak to conclude. However, now I face the similar situation: for an sequence $g_m\in C_c^\infty$, I have $g_m\to g$ strongly in $L^2$ and I have $$\lim_{n\to\infty} \int_\Omega \partial_i u_n g_m\,dx= 0 $$ Next I need to push $m\to\infty$ and I have $$\lim_{m\to\infty}\lim_{n\to\infty} \int_\Omega \partial_i u_n g_m\,dx= 0 $$ for sure. However, why could I interchange the limit here? i.e., why $$\lim_{m\to\infty}\lim_{n\to\infty} \int_\Omega \partial_i u_n g_m\,dx= \lim_{n\to\infty}\lim_{m\to\infty} \int_\Omega \partial_i u_n g_m\,dx $$

Also, I post another question regrading to how to interchange limit, but I gave a too nice condition there...

Please help, and if possible, please work out the details of how to interchange the limit. THx!!!!


1 Answer 1


If you are talking about Evans & Gariepy, "Measure Theory and Fine Properties of Functions", then either you have a different edition than I have, or your page-number is of, because my version of this book only has 256 pages (before the bibliography).

Anyway, the result which I think you are looking for is the following:

Let $(T_n)_n$ be a sequence of linear maps $T_n : X \to Y$ which are uniformly bounded in the sense that $\Vert T_n \Vert \leq C$ for some fixed $C>0$ and all $n$.

Furthermore, assume that $M \subset X$ is dense with $T_n m \to T m$ for all $m \in M$, where $T : X \to Y$ is linear and bounded.

Then $T_n x \to Tx$ holds for all $x \in X$.

For the proof, let $x \in X$ and $\varepsilon > 0$ be arbitrary. By density, there is some $m \in M$ with $\Vert x - m\Vert < \varepsilon/[3(1+C + \Vert T \Vert)]$.

For this $m$, there is some $n_0 \in \Bbb{N}$ with $\Vert T_n m - Tm \Vert < \varepsilon/3$ for all $n \geq n_0$.


\begin{eqnarray*} \Vert T_n x - Tx \Vert & \leq& \Vert T_n x - T_n m \Vert + \Vert T_n m - T m \Vert + \Vert T m - Tx \Vert \\ & \leq & C \cdot \Vert x - m\Vert + \varepsilon/3 + \Vert T \Vert \cdot \Vert x - m\Vert \leq \varepsilon \end{eqnarray*}

for all $n \geq n_0$. $\square$

To apply this in the first setting, you have to know that the sequence $(\mu_n)_n$ is bounded. In the second problem, you mention that the sequence $(u_n)_n$ is bounded in $H^1$, so that there is no problem.

In the first setting, $X = C(\Bbb{R}^n, \Bbb{R}^n)$, $Y = \Bbb{R}$ (or $\Bbb{C}$) and $M = C_c^1 (\Bbb{R}^n, \Bbb{R}^n)$, which is dense in $X$. Finally, identify each measure $\mu_n$ with the functional it induces on $X$.

In the second setting, you can take $X = H^1$, $Y = \Bbb{R}$ and $M = C_c^\infty ( \Bbb{R}^n)$. Again, identify each $u_n \in H^1$ with the linear functional it induces on $H^1$.

By the way, in the second setting, you can also apply a different (not necessarily simpler) argument: Assume that $u_n \not\to 0$ weakly in $H^1$. Then there is some $h \in H^1$ with $\langle u_n, h\rangle_{H^1} \not\to 0$, so that there is some $\varepsilon > 0$ and a subsequence $(u_{n_k})_k$ with $|\langle u_{n_k}, h\rangle_{H^1}| \geq \varepsilon$ for all $k$.

But the $H^1$ bounded sequence $(u_{n_k})_k$ has a weakly convergent subsequence $u_{n_{k_\ell}} \to v$ weakly in $H^1$. Because of $|\langle v, h\rangle_{H^1}| \leftarrow |\langle u_{n_{k_\ell}}, h\rangle_{H^1}| \geq \varepsilon$, we see $v \neq 0$.

But the embedding $H^1 \hookrightarrow L^2$ yields $u_{n_{k_\ell}} \to v$ weakly in $L^2$. But you claim that you already know $u_n \to 0$ weakly in $L^2$. By uniqueness of weak $L^2$ limits, $v = 0$, a contradiction.

  • $\begingroup$ First of all, I apologize for my mistake, I mean page 175 not 275...I already changed in my OP. $\endgroup$
    – spatially
    Oct 17, 2014 at 22:19

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