I've read the following in a few papers:

Given: Let $\Omega \subset R^d$ be a Lipschitz domain. A sequence $f_n$ converges strongly to $f$ in $L^2(0,T;L^2(\Omega))$ and weakly in $L^2(0,T;H^2(\Omega))$.

Then by a standard interpolation argument it follows that $f_n$ converges strongly to $f$ in $L^2(0,T;H^1(\Omega))$.

Now my question is what is meant by standard interpolation argument?

Thanks for your answers!

  • $\begingroup$ A comment about the title "Standard interpolation between..." -- the word "standard" in your papers refers to "argument", not to "interpolation". They are not talking about some "standard interpolation". $\endgroup$ – 40 votes Jul 17 '13 at 21:47
  • $\begingroup$ You are right, I was to quick writing the title. Any ideas what is mathematically meant by this standard argument? $\endgroup$ – Chuck Jul 18 '13 at 7:46

You may want to have a look for a book on interpolation theory, there is, e.g., a great introduction by Tartar.

There are, typically, two interpolation results: $$\|u\|_{L^p} \le \|u\|_{L^r}^\mu \, \|u\|_{L^s}^{1-\mu},$$ (Lebesgue scale), if $1/p = \mu/r + (1-\mu)/s$ and $$\|u\|_{H^1} \le C \, \|u\|_{L^2}^{1/2} \, \|u\|_{H^2}^{1/2}$$ (Sobolev scale).

Your proof requires the second one (use boundedness in $H^2$).

  • $\begingroup$ Ok thanks. So taking the square in the second interpolation inequality, integration over (0,T) and applying Hölder's inequality does the job... $\endgroup$ – Chuck Jul 20 '13 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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