Let $T>0$, $p>0$ and $f: [0,T]\to \mathbb{R}$ continuous. The $p$-variation of $f$ on $[0,T]$ is given by:$$\operatorname{Var}^p(f)_T:=\lim\limits_{n\to\infty}\sum \limits_{k=1}^{2^n}\left|f(k/2^n T)-f((k-1)/2^nT)\right|^p$$ (if the limit exsits)
I have to show that if $0< \operatorname{Var}^{p^*}(f)_T<\infty$ for a $p^*>0$, we have $\operatorname{Var}^p(f)_T=\infty$ for all $0<p<p^*$.
$0< \operatorname{Var}^{p^*}(f)_T<\infty$ for a $p^*>0$ $\Rightarrow$ $\operatorname{Var}^{p}(f)_T=0$ for all $p>p^*$.
Attempt
$f$ is continuous on a compact set, i. e. uniformly continuous.
- Let $\varepsilon >0$ such that $p=p^*+\varepsilon$. We have:
$$
\begin{split}
\sum \limits_{k=1}^{2^n}\left|f(k/2^n T)-f((k-1)/2^nT)\right|^p & =\sum \limits_{k=1}^{2^n}\left|f(k/2^n T)-f((k-1)/2^nT)\right|^{p^*+\varepsilon}\\
&\leq \max\limits_{1\leq k\leq 2^n}|f(k/2^nT)-f((k-1)/2^nT))|^{\varepsilon}\operatorname{Var}^{p^*}(f)_T.
\end{split}
$$ Since, as $n\to \infty$, we have $0< \operatorname{Var}^{p^*}(f)_T<\infty$ and $\max\limits_{1\leq k\leq 2^n}|f(k/2^nT)-f((k-1)/2^nT))|^{\varepsilon}\to 0$, we achieve the desired result.
Is this correct? (Do I need the uniform continuity for the $\leq$-part?)
- I read online that the Hölder inequality might be a good attempt, but I didn't really suceed yet. Can someone help?
For $m\in \mathbb{N}$ and a (finite) scalar sequence $(a_k)$, $\|(a_k)\|_p=\Bigl(\sum_k |a_k|^p\Bigr)^{1/p}$ if $p<\infty$, and $\|(a_k)\|_\infty = \max_k |a_k|$. Recall Hölder's inequality. If $1\leqslant \alpha,\beta\leqslant \infty$ satisfy $\alpha^{-1}+\beta^{-1}=1$, then $$\sum_k |a_kb_k|\leqslant \|(a_k)\|_\alpha \|(b_k)\|_\beta.$$ Such $\alpha,\beta$ are called conjugates or a conjugate pair.
Now fix $0<q<p<r\leqslant \infty$ and note that there exists $\theta\in (0,1)$ such that $$\frac{1}{p}=\frac{\theta}{q}+\frac{1-\theta}{r}.$$ The following interpolation inequality is pretty useful. Let $\alpha=\frac{q}{p\theta}$ and $\beta=\frac{r}{p(1-\theta)}$, so $$\frac{1}{\alpha}+\frac{1}{\beta}=p\Bigl[\frac{\theta}{q}+\frac{1-\theta}{r}\Bigr] =p[1/p]=1.$$ So $\alpha,\beta$ are conjugates. For a finite scalar sequence $(c_k)$, by applying Hölder with $\alpha,\beta$ and $a_k=|c_k|^{p\theta}$, $b_k=|c_k|^{p(1-\theta)}$,
\begin{align*} \|(c_k)\|_p & = \Bigl[\sum_k |c_k|^p\Bigr]^{1/p} = \Bigl[\sum_k |c_k|^{p\theta}|c_k|^{p(1-\theta)}\Bigr]^{1/p} \leqslant \Biggl[\Bigl[\sum_k |c_k|^{p\theta\alpha}\Bigr]^{1/\alpha} \Bigl[\sum_k |c_k|^{p(1-\theta)\beta}\Bigr]^{1/\beta}\Biggr]^{1/p} \\ & = \Bigl[\sum_k |c_k|^q\Bigr]^{1/\alpha p}\Bigl[\sum_k |c_k|^r\Bigr]^{1/\beta p} \\ & = \Bigl[\sum_k |c_k|^q\Bigr]^{\theta/q}\Bigl[\sum_k |c_k|^r\Bigr]^{(1-\theta)/r} \\ & = \|(c_k)\|_q^\theta \|(c_k)\|_r^{1-\theta}.\end{align*}
Side note number $1$: This actually says that the function $p\mapsto \|(c_k)\|_p$ is log-convex, or, equivalently, $p\mapsto \log \|(c_k)\|_p$ is convex. This is related to the Riesz-Thorin interpolation theorem and interpolation spaces in general.
Side note number $2$: This interpolation inequality is not limited to the finite scalar sequence. It's true wherever Hölder's Hölds.
Now we turn to your problem. For $n=1,2,\ldots$, let $0=i_{0,n}<i_{1,n}<\ldots < i_{2^n,n}=T$ be the partition points $Tk/2^n$, $k=0,1,\ldots,2^n$. Let $\omega(f,\delta)=\sup\{|f(x)-f(y)|:x,y\in [0,T], |x-y|\leqslant \delta\}$. As you said, $\omega(f,\delta)_\delta\to 0$ as $\delta\to 0^+$, since $f$ is uniformly continuous. For each $n\in \mathbb{N}$ and $0<p<\infty$, let $$H_{n,p}(f)=\Bigl[\sum_{k=1}^{2^n} |f(i_{k,n})-f(i_{k-1,n})|^p\Bigr]^{1/p} = \|(|f(i_{k,n})-f(i_{k-1,n})|)_{k=1}^{2^n}\|_p=.$$ Note that, if the limit exists, $$\lim_n H_{n,p}(f)=[\text{Var}^p(f)_T]^{1/p}.$$
Fix $0<q<p$ and let $r=\infty$. With $\theta=q/p\in(0,1)$, we can apply the interpolation inequality above to find that $$H_{n,p}(f)\leqslant H_{n,q}(f)^\theta\cdot [\max_k |f(i_{k,n})-f(i_{k-1,n})|]^{1-\theta} \leqslant H_{n,q}(f)^\theta \cdot \omega(f,2^{-n})^{1-\theta}.$$
On one hand, if we assume that the limit in the definition of $\text{Var}^q(f)_T$ exists and is finite, then we have $$\underset{n}{\lim\sup} H_{n,p}(f) \leqslant \lim_n H_{n,q}(f)^\theta \lim_n \omega(f,2^{-n})^{1-\theta} = [\text{Var}^q(f)_T]^{\theta/q}\cdot 0=0.$$ Thus the $\lim\sup$ is actually a limit. Raising both sides to the power $p$ gives $\text{Var}^p(f)_T=\lim_n H_{n,p}(f)^p=0$.
On the other hand, if we assume that the limit in the definition of $\text{Var}^p(f)_T$ exists and is positive, then we have $$\underset{n}{\lim\inf} H_{n,q}(f) \geqslant \Bigl[\frac{\lim_n H_{n,p}(f)}{\omega(f,2^{-n})^{1-\theta}}\Bigr]^{1/\theta}.$$ Since the numerator tends to a finite value and the denominator goes to $0$, we have that the $\lim\inf$ is actually $\infty$, and the "limit" is $\infty$. Here it is worth noting that we are not dividing by zero. That's because if $\omega(f,2^{-n})=0$, then $f$ is actually constant, which is not possible, since we have assumed that $\text{Var}^p(f)_T>0$. Now, raising both sides to the power $q$, we get $\text{Var}^q(f)_T=\lim_n H_{n,q}(f)^q=\infty$.