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nivel
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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?

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?

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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nivel
  • 45
  • 3

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?