MathOverflow is a question and answer site for professional mathematicians. 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

We know that Morrey's inequality says $W^{1,p} \subset C^{0,\gamma}$ for $\gamma = 1 - n/p$ where $n$ is the dimension. However, in 1D, following the proof of Evans "Partial Differential Equations" (first edition, pp. 267), we have for any function $f$ in $W^{1,p}(0,1)$, and $x, y \in (0,1)$

$u(x) - u(y) = \int_0^1 \frac{d}{dt}u(tx + (1-t)y) dt = \int_y^x Du(s) (x-y) ds$.

Take absolute value on both sides and use the basic inequalities, we have

$|u(x)-u(y)| \le |Du|_{L^p(0,1)} |x-y|$.

We obtain something more than Morrey's inequality would indicate.

Is there anything wrong?

share|cite|improve this question
What basic inequalities are you using? maybe you should check again the statement of Holder's inequality... – Piero D'Ancona Oct 3 '10 at 0:21
I read this too quickly. You should show the details of your argument. But you should do this on – Deane Yang Oct 3 '10 at 1:03
Yes. I there is a change of variable mistake in the details I did not show (in the second equality). Thanks. – nivel Oct 3 '10 at 2:38
up vote 5 down vote accepted

This does not look right. You have $u(x)-u(y) = \int_y^x Du(s)(x-y)ds$ but the correct expression is clearly $\int_y^x Du(s)ds$. I think a change of variables went wrong somewhere.

A counterexample: take $u(x) = x^{3/4}$ on $(0,1)$, with $p=2$. Then $Du(x) = \frac{3}{4}x^{-1/4} \in L^2(0,1)$, so $u \in W^{1,2}(0,1)$, but $u$ is not Lipschitz.

share|cite|improve this answer
I see, thanks. When I changed the variable to s, I need the $x-y$ back. Thanks. – nivel Oct 3 '10 at 2:09

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.