Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $g_k,g\in H^1(\Omega)$ (bounded domain) be given, with $g_k\to g$ in $L^2(\Omega)$. Unfortunately, I don't know whether the $g_k$ are uniformly bounded in $H^1$. I want to show that $g_k\rightharpoonup g$ in $H^1(\Omega)$. But this seems to be intricate if not impossible.

However, with the strong $L^2$ convergence I am able to prove that

$$\partial_j g_k \to \partial_j g\text{ in } \mathcal{D}'(\Omega)$$

Can one deduce from that (or with a different approach) that the convergence is weak in $L^2(\Omega)$?

I fear that this is not possible since one needs to put the derivatives on the test function to prove the distributional convergence.

share|cite|improve this question
up vote 2 down vote accepted

Distributional convergence is very weak indeed, because it is implied by $L^1$ convergence, and passes to derivatives of all orders. For example, since $n^{-1}\sin n^2x\to 0$ in $L^1(0,\pi)$ (and in $L^2(0,\pi)$ as well), it follows that $n\cos nx\to 0$ in the sense of distributions. But the sequence $n\cos nx$ is not bounded in $L^2$, hence cannot converge weakly. Hence, $n^{-1}\sin n^2x$ does not converge weakly in $H^1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.