Is $\int_0^\infty \sin(Kx)f_K(x)\,dx$ of larger order than $\int_0^\infty \cos(Kx)f_K(x)\,dx$?

Question

Suppose we have a function $f$, such that $f$ is of some smoothness degree $m$, and $f,f^{(k)} \in L_1[0,\infty)$ $k=1,...,m$. Now if $f^{(k)}(0) = \lim_{x\rightarrow\infty}f^{(k)}(x) = 0$ for $k=1,...,m-1$ then we can impose the standard bound on the decay of the Sine and Cosine transforms to be $O(K^{m-1})$ where the Sine and Cosine transforms are $\int_0^\infty \sin(Kx)f(x)dx$ and $\int_0^\infty \cos(Kx)f(x)dx$ respectively. If, however $f(x)\neq 0$ we can say something different. In fact we would be able to say that the Sine transform decays slower than the Cosine transform at least by a factor of $K^{-1}$. The mathematical reasoning behind this of course lies within the Riemann Lebesgue Lemma, and Integration by parts:

$$ \begin{align} \int_0^\infty\sin(Kx)f(x)dx & = \left[-K^{-1}\cos[Kx]f(x)\right]_0^\infty + K^{-1}\int_0^\infty \cos(Kx)f^\prime(x)dx\\ & = K^{-1}f(0) + K^{-1}o(1)\\ & = K^{-1}f(0)(1+o(1)). \end{align} $$

If we then consider the Cosine transform, we have that $$ \begin{align} \int_0^\infty\cos(Kx)f(x)dx & = K^{-1}o(1) = o\left(\left|\int_0^\infty \sin(Kx)f(x)dx\right|\right), \end{align} $$

and so the cosine transform decays faster than the sin transform does in magnitude. The intuition behind this is that the decaying property for the Sine and Cosine transform depends on the behaviour of our function $f$ at zero. The sine transform can't deal with the jump at the origin because all Sine waves start at the origin so as $K$ becomes larger, you 'need more' Sine waves to try and capture that jump at zero, whereas since the Cosine transform is able to capture it because the Cosine function has a 'rescaled' jump at the origin itself as well.

QUESTION: My question concerns itself with an extension of this property. Suppose now that I have a sequence of functions $f_K(x)$ belonging in the same class as $f(x)$ as above, with $f_K(0) \neq 0$ but $f_K(0)\rightarrow 0$. Can I still suggest something like $\int_0^\infty \cos(Kx)f_K(x)dx = o\left(\left|\sin(Kx)f_K(x)\right|\right)$? My intuition says yes since at each step $K$, we always have a jump discontinuity at the origin and the Sine function cannot deal with this, and so you will always need more Sine functions of frequency ~ K than you would of Cosine functions. If you do integration by parts, the question is then similar to that of trying to show that

$$\begin{align} \int_0^\infty\cos(Kx)f_K^\prime(x)dx = o(f_K(0)), \end{align} $$

and it should be enough to have the cosine transform bounded above at least by some constant multiple of the Sine transform. Any ideas would be appreciated.

Answer 1

In general, clearly no.

Let $f$ be a fixed function of your class with $f(0) = 1$. Let $f_K(x) = \frac{1}{K}f(Kx)$. Then $f_K$ is an admissible sequence by your hypothesis. However

$$\int_0^\infty \cos(Kx) f_K(x) \mathrm{d}x = \int_0^\infty \frac{1}{K^2} \cos(y) f(y) \mathrm{d}y $$

and similarly

$$\int_0^\infty \sin(Kx) f_K(x) \mathrm{d}x = \int_0^\infty \frac{1}{K^2} \sin(y) f(y) \mathrm{d}y $$

so both sequences decay exactly at the rate $\frac{1}{K^2}$, provided they are non-zero. If you choose $f$ such that $\int_0^\infty \cos(x) f(x) \mathrm{d}x \neq 0$ but $\int_0^\infty\sin(x) f(x) \mathrm{d}x = 0$, then the second sequence is always 0 and your comparison principle fails again.

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Basically, Riemann-Lebesgue gives you no rate. And when you have no rates, the diagonal of a double sequence can behave very differently from the horizontal or vertical parts.