1
$\begingroup$

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.

| cite | improve this question | |
$\endgroup$
1
$\begingroup$

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.

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That however is an example of a case where the functions decays uniformly to zero everywhere on the domain. What happens when you are looking for something that doesn't go to zero on an unbounded lebesgue measurable set? So in that instance, I think it makes sense that both decay at the same rate, seeing as the resulting function is identically zero..? what about cases when this is not true? $\endgroup$ – user61038 Mar 23 '14 at 23:52
  • $\begingroup$ I'm sorry but I cannot understand at all what you are trying to ask in your comment. Can you be a bit more precise? $\endgroup$ – Willie Wong Mar 24 '14 at 8:25
  • $\begingroup$ In your above example, you use an example $f_k(x) = \frac{1}{K} f(Kx)$ which by my above hypothesis fits the class of functions that I am looking at. However, this example is a specific example of functions which decay to zero on the whole real line. Precisely, $\|(1/K)f(Kx)\| \leq (1/K)\|f\| \rightarrow 0$ for all $x$. So I believe in some intuitive sense, the Sine transform and Cosine transform should decay at the same rate, because the limiting function is identically zero everywhere, and more specifically zero uniformly everywhere. $\endgroup$ – user61038 Mar 25 '14 at 0:39
  • $\begingroup$ I suppose I should have been a bit more specific in my question above, but what I am more concerned with are functions $f_K$ such that if $f_K\rightarrow f_\infty$, $f_\infty$ is non-trivial i.e. $f_\infty(x) \neq 0$ for all $x$, but $f_K(0)\rightarrow f_\infty(0)=0$. I feel in this case, there are certain restrictions that can be imposed on $f$ such that either the Sine transform or Cosine transform decays faster than the other. $\endgroup$ – user61038 Mar 25 '14 at 0:39
  • $\begingroup$ In regards to your last comment: let $g$ be a fixed, non trivial $C^\infty$ function of compact support on $(0,\infty)$. We have that $\hat{g}$ decays arbitrarily fast since it is Schwartz. Consider the sequence of functions $f_K + g$. Since $\hat{g}$ is Schwartz, its contribution to the sine and cosine transforms are eventually negligible. $f_K + g \to g$ as $K\to \infty$ in any $L^p$ with $p \geq 1$, the limit is clearly non-trivial. (For any linear operation changing the [strong] limit will not get you any different result.) $\endgroup$ – Willie Wong Mar 25 '14 at 10:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.