How to analyze the following complex number?

Question

$f(T) = \int_{0}^{1}e^{-iT\lambda(s)}ds$, where $\lambda(s)$ increases monotonously. I want to show that $|f(T)|$ has an upper bound that decreases with $T$ for $T \gg 1$.

For example, if $\lambda(s) = s$ then $|f(T)| \le 2/T$.

This question comes from numerical analysis of my research. It's the expression of error analysis in adiabatic evolution. More specifically, the digital error has expression : $\sum_{k=1}^{L}e^{-iT\sum_{j<k}\lambda_{j}}/L$, where $\lambda_{j}$ equals $\lambda_{j} = \lambda(j/L)$. From numerical result I am sure it has an upper which bound decays with $T$. However, I don't know how to prove that. I wonder if there is related formulas in complex analysis and fourier series.

Answer 1

Remark: This answer was for the first version of the question which stated that $|f(T)|^2$ was monotonic.

The result is not true. If $\lambda(s) = s$, then \begin{equation} f(T) = \frac{1}{i T}(1 - e^{-i T}) \end{equation} hence $f(2k\pi) = 0$ for $k\in {\mathbb{N}^*}$ but $f$ is not identically $0$, hence $|f(T)|^2$ is not monotonic.

Answer 2

I make here two assumptions on $\lambda$

1. it has derivative bounded from below by m > 0
2. derivative is monotonous (so $\frac{1}{\lambda^{'}}$ is also monotonous)

First off, if f(t) is a smooth function then $$f(s_{k}) - f(s_{k-1}) = f^{'}(s_{k-1})*(s_{k}-s_{k-1}) + \epsilon(s_{k}-s_{k-1})$$ This is just a mean value theorem. Here $s_{1} < s_{2} < s_{3} <... <s_{n}$ is a partition of [0,1] and $\epsilon$ stands for something which tends to zero (uniformly) when $s_{k} - s_{k-1}$ tends to zero. Using this for $f(s) = e^{iT\lambda(s)}$ one gets $$e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}= iT\lambda^{'}(s_{k-1})e^{iT\lambda(s_{k-1})}(s_{k}-s_{k-1}) + \epsilon*(s_{k}-s_{k-1})$$ so $$\frac{1}{iT\lambda^{'}(s_{k-1})}e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})} = e^{iT\lambda(s_{k-1})}(s_{k}-s_{k-1}) + \frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}$$ Now, writing the integral $\int_{0}^{1}e^{iT\lambda(s)}ds$ as the limit of Riemann' sums we have $$\int_{0}^{1}e^{iT\lambda(s)}ds = \lim_{\Delta s \to 0}\sum_{k}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})})(s_{k}-s_{k-1})$$ But from the formula above we see that the sum can be estimated by $$\sum_{k}\frac{1}{iT\lambda^{'}(s_{k-1})}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}) - \sum_{k}\frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}$$ The second sum is easy to bound from above (i.e. assuming that $\lambda^{'}$ is bound from below by, say m) we get $$|\sum_{k}\frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}| <= $$ $$\sum_{k}|\frac{\epsilon*(s_{k}-s_{k-1})}{iT\lambda^{'}(s_{k-1})}| <= $$ $$\frac{1}{Tm}\sum_{k}|\epsilon*(s_{k}-s_{k-1})| <=$$ $$\frac{max(\epsilon)}{Tm}\sum_{k}|(s_{k}-s_{k-1})| <=$$ $$\frac{max(\epsilon)}{Tm}$$ Now we have to estimate the first sum $$\frac{1}{iT}\sum_{k}\frac{1}{\lambda^{'}(s_{k-1})}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}) $$ When we change order of summation we get that the sum is $$|\frac{1}{iT}\sum_{k}\frac{1}{\lambda^{'}(s_{k-1})}(e^{iT\lambda(s_{k})} - e^{iT\lambda(s_{k-1})}) | <= $$ $$|\frac{1}{iT}\sum_{k}(\frac{1}{\lambda^{'}(s_{k})} - \frac{1}{\lambda^{'}(s_{k-1})}) e^{iT\lambda(s_{k-1}}) | + 2\frac{1}{Tm} <= $$ $$\frac{1}{T}\sum_{k}|(\frac{1}{\lambda^{'}(s_{k})} - \frac{1}{\lambda^{'}(s_{k-1})}) e^{iT\lambda(s_{k-1}}) | + 2\frac{1}{Tm} <= $$ $$\frac{1}{T}\sum_{k}|(\frac{1}{\lambda^{'}(s_{k})} - \frac{1}{\lambda^{'}(s_{k-1})}) | + 2\frac{1}{Tm} <= $$ And since $\frac{1}{\lambda^{'}}$ is by assumption monotone we can skip absolute values under the last sum to get $$\frac{1}{T}|\frac{1}{\lambda^{'}(0)} - \frac{1}{\lambda^{'}(1)} | + 2\frac{1}{Tm} $$

Answer 3

It suffices to prove that $f(T)$ tends to $0$ when $T$ tends to infinity, because this is equivalent to having an upper bound that decreases to $0$, namely $U(T) = \sup_{T'\ge T}|f(T')|$.

Let us suppose that $\lambda(s)$ is differentiable and that $\lambda'(s)$ never vanishes. The substitution $x = \lambda(s)$ gives \begin{equation} f(T) = \int_{\lambda^{-1}([0,1])}e^{-iTx}\frac{1}{|\lambda'(\lambda^{-1}(x))|} d x \end{equation} The Riemann-Lebesgue lemma now tells us that $\lim_{T\to\infty}f(T) = 0$ because $\chi_{x\in\lambda^{-1}([0,1])} \frac{1}{|\lambda'(\lambda^{-1}(x))|}\in L^1({\mathbb R})$. Indeed, its integral is $f(0) = 1$.

I'm pretty confident the differentiability assumption can be replaced by a weaker condition with some work.