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

I am quite new to the field of Sobolev spaces. So, I want to apologize in advance, if the question is too obvious!

I have a problem with understanding the connection between the Hilbert space $H^2(0,1)$ and $C^2[0,1]$. I know, that due to the Sobolev Embedding theorem the embedding $E: H^2(0,1)\to C[0,1] (\text{and} C^1[0,1])$ is compact. But what could be said about $C^2[0,1]$? Can we say, that $C^2[0,1]\subset H^2(0,1)$?

I would really appreciate the answer.

share|cite|improve this question
Wouldn't you have even $H^1(0,1)\subset C[0,1]$? – timur Sep 27 '12 at 11:37
Yes, due to the Sobolev Embedding Theorem. – Dina Sep 28 '12 at 11:24
Would you even have $W^{1,1}(0,1)\subset C[0,1]$? – timur Sep 28 '12 at 14:29
up vote 3 down vote accepted

Yes, because you have a bound $$\int_0^1\lvert f^{(k)}(x)\rvert^2\,dx\le \lVert f^{(k)}\rVert_\infty^2\qquad\text{for }k=0,1,2,$$ thus bounding the $H^2$ norm of $f$ in terms of the $C^2$ norm. Here $\lVert\cdot\rVert_\infty$ is the sup norm.

share|cite|improve this answer
Thank you, that was really easy! – Dina Sep 27 '12 at 11:28

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.