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I have a question about p-variation.

I'm reading A first Course in Sobolev Spaces by Giovanni Leoni. In this book, a confused construction is demanded (Exercise 2.26):

Let $p\ge1$.Given a function $u:[a,b]\rightarrow\mathbb{R}$ for every $\varepsilon>0$ let $$Var_pu=:sup\{(\sum_{i=1}^n|u(x_i)-u(x_{i-1})|^p)^{1/p}\}$$ where the supremum is taken over all partitions $P=\{x_0,\ldots,x_n\}$ of $I$ and $$Var^{(\varepsilon)}_{p}u:=sup\{(\sum_{i=1}^{n}|u(x_i)-u(x_{i-1})|^p)^{1/p}\},$$ where the supremum is taken over all partitions $P=\{x_0,\ldots,x_n\}$ of $I$ such that $0<x_i-x_{i-1}\le\varepsilon,i=1,\ldots,n, n\in\mathbb{N}$. Define $$Var^{(0)}_{p}u:=\lim_{\varepsilon\rightarrow0^{+}}Var^{(\varepsilon)}_{p}u.$$ (i) Prove that $Var^{(0)}_{p}u<\infty$ if and only if $Var_{p}u<\infty$.

(ii) Prove that there exists a function $u$ for which $Var^{(0)}_{p}u<Var_{p}u<\infty$

My Attempt:

It's easy to find $Var^{(0)}_{p}u\le Var^{(\varepsilon)}_{p}u\le Var_{p}u\;$for every $\varepsilon$ small enough. And I want to note that $Var^{(0)}_{p}u\ge Var_{p}u$ in the following way.

Fix $\delta>0$, we can find a $\varepsilon>0$ such that $Var^{(\varepsilon)}_{p}u-Var^{(0)}_pu\le\delta$ by the definition of $Var^{(0)}_pu$. And there exists some partition $P=\{x_0,\ldots,x_n\}$ of $[a,b]$ that satisfies $$Var_pu-\{\sum^{n}_{i=1}|u(x_i)-u(x_{i-1})|^p\}^{1/p}\le\delta,i=1,\ldots,n,n\in\mathbb{N}$$

Now we consider a more detailed partition $P^{'}=\{x^{'}_0,\ldots,x^{'}_{n}\}$ of $[a,b]$ which is generated by adding the k-section points in every $[x_{i-1},x_i]$. $P^{'}$ will be the $\varepsilon$-partition if we choose the integer $k$ big enough. The Minkowski inequality guarantees the following inequality: $$Var_pu\le\{\sum^{n}_{i=1}|u(x_i)-u(x_{i-1})|^p\}^{1/p}+\delta\le\{\sum^{n}_{i=1}|u(x^{'}_i)-u(x^{'}_{i-1})|^p\}^{1/p}+\delta\le Var^{(\varepsilon)}_pu+\delta\le Var^{(0)}_pu+2\delta$$ Finally, let $\delta\rightarrow0^+$ and we get $Var_pu=Var^{(0)}_pu$ ???

It may be a ridiculous result but I can't find the possible error below. I would like to know the solution to this exercise or the error in my proof. And if you know references and books related to this exercise, please let me know.

Thanks in advance.

1 Answers1

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Your result is false. Take the path $X(t) = t$ on $ [0,1]$. It has got $2$-variation equal to $1$, but the $2$-variation with a supremum taken over partitions with mesh size tending to zero (i.e., your $V^{(0)}$), is zero, by convexity.