# In 1D, is a $W^{1,p}$ function always Lipschitz, for $p\ge1$?

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?

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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 math.stackexchange.com –  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

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