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Mar 26 '14 at 0:07 comment added user61038 This is a very insightful response. Thankyou for your time and effort.
Mar 25 '14 at 10:07 comment added Willie Wong The above is in the case where $m \geq 2$. If $m = 1$ you can construct a $C^1$ function $g$ with compact support such that its Fourier transform decays no faster than $|\xi|^{-1-\epsilon}$. In fact you can make it such that for some sequence of integers $K_1, \ldots$ both the sine and cosine transform are exactly of size $|K_i|^{-1 - \epsilon}$. Then looking at $f_K + g$ the two sequences you want would both eventually be dominated by this slower decay and again, you cannot say that one of the sine or cosine transforms decay faster.
Mar 25 '14 at 10:03 comment added Willie Wong 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.)
Mar 25 '14 at 0:39 comment added user61038 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.
Mar 25 '14 at 0:39 comment added user61038 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.
Mar 24 '14 at 8:25 comment added Willie Wong 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?
Mar 23 '14 at 23:52 comment added user61038 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?
Mar 21 '14 at 12:26 history answered Willie Wong CC BY-SA 3.0