The derivative of a moving L2 norm

Question

Consider the class of $C^0$ signals $x(t)$ such that $$ \left|x\left(t+T\right)-x\left(t\right)\right|\leq\epsilon $$ I would like to find a uniform bound on the derivative of its moving rms, which is defined by $$ J\left(t\right)=\left(\frac{1}{T}\int_{t-T}^{t}\left|x\left(\tau\right)\right|^{p}d\tau\right)^{\frac{1}{p}} $$ with $p=2$. Apparently, when $\epsilon=0$, $J$ is always constant and the derivative is uniformly 0; so I wonder whether $ \left|\frac{dJ}{dt}\right| $ is uniformly small when $\epsilon$ is small.

It might be worth noting that a uniform bound could be found when $p=1$ (in the case of moving average), see https://math.stackexchange.com/q/57307.

It would also be interesting to find out whether it holds for other $p\in[1,\infty]$.

Answer 1

I will consider the case $p=2$ and (for convenience) the slightly different RMS definition $$ J(t) = \left(\frac{1}{T}\int_{t-T}^t|x(\tau)|^2 d\tau\right)^{1/2}. $$

Let $K(t)\triangleq J^2(t)$. Assuming that $x(t)$ is continuous, it follows from the Leibniz integral rule that $$ \frac{dK}{dt} = \frac{1}{T}\left(|x(t)|^2-|x(t-T)|^2 \right) $$ Since it also holds $\frac{dK}{dT}=2J\frac{dJ}{dt}$, we have an explicit expression for the derivative of $J(t)$, $$ \begin{align} \frac{dJ}{dt}&=\frac{1}{2TJ(t)}\left(|x(t)|^2-|x(t-T)|^2 \right)\\ &=\frac{1}{2TJ(t)}\left(|x(t)|-|x(t-T)| \right) \left(|x(t)|+|x(t+T)| \right) \end{align} $$ Taking the absolute value on both sides and recalling that $|x-y|\geq ||x|-|y||$, a bound for the magnitude of the derivative can be obtained as $$ \left|\frac{dJ}{dt}\right| \leq \frac{|x(t)|+|x(t-T)| }{2T} \frac{|x(t)-x(t+T)|}{|J(t)|} $$ Noting that

• The first ratio of the RHS is bounded (continuity of $x(t)$),
• $J(t)\geq \frac{1}{T}\int_{t-T}^{t}|x(\tau)|d\tau$ (power means inequality, see, e.g., [The Cauchy-Schwarz Master Class, Ch. 8, pp. 127-128]) with the RHS tending to a constant as $\epsilon\rightarrow 0$,

it follows that $$ \left|\frac{dJ}{dt}\right| \rightarrow 0 $$ as $\epsilon \rightarrow 0$, implying that $\left|\frac{dJ}{dt}\right|$ is small when $\epsilon$ is small.

Comment: The above hold under the assumption $J(t)\neq 0$. However, $J(t)=0$ if and only if $x(t)=0$ for all $t$ (continuous $x(t)$), which is a case of no practical interest.

Answer 2

I think I've found a counterexample for such a uniform bound to exist. Consider the following sequence of functions: $$ x_{k}\left(t\right)=\begin{cases} \epsilon n\cos\left(\omega_{k}t\right) & t\in\left[2\pi n+\pi-\frac{\pi}{2\omega_{k}},2\pi n+\pi+\frac{\pi}{2\omega_{k}}\right]\\ 0 & \text{otherwise} \end{cases} $$ where $k$ is the index, $\omega_{k}=k$ and $n\left(t\right)$ is a step function which increments its value upon every cycle of $T=2\pi$. The graph of each function essentially looks like a bunch of pinnacles rising above the level plane with increasing magnitude. As the index $k$ grows, the pinnacles become thinner. It can be shown that $$ \left\Vert x_{k}\left(t+T\right)-x_{k}\left(t\right)\right\Vert _{\infty}=\epsilon $$ for all $t\geq0$ and $k\in\mathbb{N}^{+}$. However, the derivative grows unbounded as $k\rightarrow\infty$. The figure below shows the $k=1$ and $k=5$ cases.

This result may not be that surprising from @Stelios's answer, noting the fact that $$ \frac{dJ}{dt}=\frac{1}{2TJ(t)}\left(|x(t)|-|x(t-T)|\right)\left(|x(t)|+|x(t+T)|\right) $$ In this particular counterexample, the value of $\frac{1}{2T}$, $\left(|x(t)|-|x(t-T)|\right)$ and $\left(|x(t)|+|x(t+T)|\right)$ are the same at corresponding peaks, but the value $J(t)$ at these peaks tend to zero as $k$ grows, so the derivative also goes unbounded. Therefore, sorry about asking this question, but I think there may not be a uniform bound unless further assumptions are imposed (something to do with how spread-out the functions are).