If $f\in C([a,b])$, $V(f^2;[a,b])<\infty$, then $V_2(f;[a,b]) \leq C(V(f^2;[a,b])+\|f\|_\infty^2)$

Question

Assume that $f\in C([a,b])$ is a complex-valued function with $V(f^2;[a,b])<\infty$. Prove that $$V_2(f;[a,b])\leq C(V(f^2;[a,b])+\|f\|_\infty^2),$$ where $C>0$ is independent of $a,b$ and $f$.

Note that $V(f;[a,b])$ denotes the total variation of $f$ on $[a,b]$: $$V(f;[a,b]):=\sup_{a=a_0<a_1<\cdots a_k=b}\sum_{j=1}^k|f(a_j)-f(a_{j-1})|.$$ $V_2(f;[a,b])$ denotes the quadratic variation of $f$ on $[a,b]$: $$V_2(f;[a,b]):=\sup_{a=a_0<a_1<\cdots a_k=b}\sum_{j=1}^k|f(a_j)-f(a_{j-1})|^2.$$

I can only prove the inequality $V_2(f;[a,b])\leq 2V(f^2;[a,b])$ for real-valued $f$, by first showing $V_2(f;[a,b])\leq 2V_2(|f|;[a,b])$, thanks to the continuity of $f$, and then showing $V_2(|f|;[a,b])\leq V(f^2;[a,b])$. Both steps are consequences of the definitions and require the function to be real-valued. In the first step, the IVT (for real-valued functions) is used; and in the second step, I use the fact that $\sqrt x$ is $1/2-$H$\ddot{\text{o}}$lder continuous: $|\sqrt x-\sqrt y|\leq\sqrt{|x-y|}$ for $x,y\geq 0$.

Here is my detailed proof of real-valued case.

Step1. Let $a=a_0<a_1<\cdots<a_k=b$. Consider $\sum_{j=1}^k|f(a_j)-f(a_{j-1})|^2$. For each $1\leq j\leq k$, if $f(a_j)\cdot f(a_{j-1})\geq 0$, then $|f(a_j)-f(a_{j-1})|=\left||f(a_j)|-|f(a_{j-1})|\right|$; if $f(a_j)\cdot f(a_{j-1})<0$, by IVT, there exists $\xi_j\in(a_{j-1}-a_j)$ such that $f(\xi_j)=0$, so

\begin{align*} |f(a_j)-f(a_{j-1})|^2&\leq(|f(a_j)-f(\xi_j)|+|f(\xi_j)-f(a_{j-1})|)^2\\ &\leq2|f(a_j)-f(\xi_j)|^2+2|f(\xi_j)-f(a_{j-1})|^2\\ &=2||f(a_j)|-|f(\xi_j)||^2+2||f(\xi_j)|-|f(a_{j-1})||^2. \end{align*}

Therefore, we have $$\sum_{j=1}^k|f(a_j)-f(a_{j-1})|^2\leq 2 V_2(|f|;[a,b]).$$ Taking supremum gives that $V_2(f;[a,b])\leq 2V_2(|f|;[a,b])$.

Step2. Let $a=a_0<a_1<\cdots<a_k=b$. Since $|\sqrt x-\sqrt y|\leq\sqrt{|x-y|}$ for $x,y\geq 0$, we have

\begin{align*} \sum_{j=1}^k||f(a_j)|-|f(a_{j-1})||^2&=\sum_{j=1}^k\left|\sqrt{f^2(a_j)}-\sqrt{f^2(a_{j-1})}\right|^2\\ &\leq\sum_{j=1}^k|f^2(a_j)-f^2(a_{j-1})|\leq V(f^2;[a,b]). \end{align*}

So $V_2(|f|;[a,b])\leq V(f^2;[a,b])$.

Combining Steps 1 and 2 gives $V_2(f;[a,b])\leq 2V(f^2;[a,b])$ for real-valued $f$.

However, I don't know how to deal with the complex-valued case. I tried to write $f$ in the form of $f=f_r+if_i$, where $f_r$ and $f_i$ are real-valued, and then followed the same ideas as in the proof of real-valued case. But things went to be messy.

This question relates to another question I asked before. I want to prove the theorem mentioned in that post:

Theorem. If $m\in C(\mathbb R)$ is bouneded such that $m^2$ has bounded variation, then $m$ is an $L^p(\mathbb R)$ Fourier multiplier.

I can prove the theorem for $m$ with $m\in V_2$. Therefore, to finish the proof of the theorem, I only need to show that if $m^2$ is of bounded variation, then $m$ is of bounded quadratic variation, which is exactly this post deals with.

Since in many circumstances, we have to consider comlex-valued multipliers, for example, differential operators. So I want to know the proof of complex-valued functions.

Any help would be appreciated!

Answer 1

This is to deal with issues related to the choice of argument for $f$ when $f$ has zeroes in $[a,b]$.

Let $a=t_0<\cdots< t_n=b$ be any partition of $[a,b]$ and consider the sum $$\sum^n_{j=1}|f(t_j)-f(t_{j-1})|^2$$

• If $f(t_j)f(t_{j-1})=0$, then $$|f(t_j)-f(t_{j-1})|^2=|f^2(t_j)-f^2(t_{j-1})|$$
• If $f(t_j)f(t_{j-1})\neq0$ and $f$ has a zero $\xi_j\in(t_{j-1},t_j)$, then $$|f(t_j)-f(t_{j-1})|^2\leq 2\big(|f^2(t_j)-f^2(\xi_j)|+|f^2(\xi_j)-f^2(t_{j-1})|\big)$$
• Finally, if $f(t_j)f(t_{j-1})\neq0$ and $f$ has no zero in $(t_{j-1},t_j)$, then there is a continuous choice of argument of $f$ in $[t_{j-1},t_j]$, that is, there is a continuous function $\theta:[t_{j-1},t_j]\rightarrow\mathbb{R}$ such that $$\frac{f(t)}{|f(t)|}=e^{i\theta(t)}\qquad\forall t\in[t_{j-1},t_j]$$ See this posting, for example, which discusses continuous branches of logarithm and argument, or sections 5.4 and 7.2 of Beardon, A. F., Complex Analysis: The Argument Principle in Analysis and Topology, John Wiley & Sons. New York, 1979. \
  • If $\Delta_j\theta=\theta(t_j)-\theta(t_{j-1})\in[0,\pi/2]\cup[3\pi/2,2\pi)\mod 2\pi$, then the OP's geometric argument in the posting yields $$|f(t_j)-f(t_{j-1})|^2\leq |f^2(t_j)-f^2(t_{j-1})|$$
  • If $\Delta_j\theta\in(\pi/2,3\pi/2)\mod2\pi$, then by continuity of $\theta$, there are $\xi_1,\xi_2\in(t_{j-1},t_j)$ such that $\theta(t_j)-\theta(\xi_2)$, $\theta(\xi_2)-\theta(x_1)$, and $\theta(x_1)-\theta(t_{j-1})$ are in $[0,\pi/2]\cup[3 p/2,2\pi)\mod 2\pi$. Then, $$\begin{align}|f(t_j)-f(t_{j-1})|^2&\leq \big(|f(t_j)-f(\xi_2)|+|f(\xi_2)-f(\xi_1)|+|f(\xi_1)-f(t_{j-1})|\big)^2\\ &\leq3\Big(|f(t_j)-f(\xi_2)|^2+|f(\xi_2)-f(\xi_1)|^2+|f(\xi_1)-f(t_{j-1})|^2\Big)\\ &\leq3\Big(|f^2(t_j)-f^2(\xi_1)|+|f^2(\xi_2)-f^2(\xi_1)|+|f^2(\xi_1)-f^2(t_{j-1})|\Big) \end{align} $$ by the same geometric argument.

Putting things together, we obtained that $$V_2(f;[a,b])\leq 3V(f^2;[a,b])$$

Answer 2

Now I know how to prove the inequality for complex-valued $f$. Reading again my proof for real-valued $f$ tells me that what I've proved is: for any $-\infty<x<y<+\infty$ we can find $\xi\in[x,y]$ such that $$|f(x)-f(y)|^2\leq 2\Big(|f^2(x)-f^2(\xi)|+|f^2(\xi)-f^2(y)|\Big). \tag{1}$$ Indeed, inequality $(1)$ also holds for complex-valued functions $f$.

Assume that $f(x)=r_1e^{i\theta_1}$, $f(y)=r_2e^{i\theta_2}$ and $\theta:=\theta_2-\theta_1\in[-\pi,\pi]$.

If $|\theta|\leq \frac\pi2$, then we can easily check that $|f(x)-f(y)|\leq |f(x)+f(y)|$ (see the picture below for a quick proof), so $$|f(x)-f(y)|^2\leq |f(x)-f(y)||f(x)+f(y)|=|f^2(x)-f^2(y)|.$$ Taking $\xi=x$ we get $(1)$.

If $\frac\pi2<|\theta|\leq\pi$, then by the continuity of $f$, there exists $\xi\in[x,y]$ such that $$f(\xi)\in\left\{re^{i\frac{\theta_1+\theta_2}{2}}: r\in[0,\infty)\right\}.$$ Since $\left|\theta_1-\frac{\theta_1+\theta_2}{2}\right|=\frac{|\theta|}2\leq\frac\pi2$, as before we have $|f(x)-f(\xi)|^2\leq |f^2(x)-f^2(\xi)|$; the same reason implies that $|f(y)-f(\xi)|^2\leq |f^2(y)-f^2(\xi)|.$ So \begin{align*} |f(x)-f(y)|^2&\leq 2\Big(|f(x)-f(\xi)|^2+|f(\xi)-f(y)|^2\Big)\\ &\leq 2\Big(|f^2(x)-f^2(\xi)|+|f^2(\xi)-f^2(y)|\Big). \end{align*} We also get $(1)$.

Finally, let me restate what we have proved: Assume that $f\in C([a,b])$ is a complex-valued function with $V(f^2;[a,b])<\infty$. Then $$V_2(f;[a,b])\leq 2V(f^2;[a,b]).$$