Timeline for An Integral Functional Equation
Current License: CC BY-SA 3.0
17 events
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| Oct 21, 2013 at 8:48 | comment | added | svangen | Gerarld Edgar: yes of course it is a bit hand wavy. Whenever e.g. $c[\delta_a]$ is mentioned, one should really consider the limit of $c[p_n]$, where $p_n$ is nonnegative and integrable and approximates $\delta_a$. | |
| Oct 18, 2013 at 16:25 | comment | added | Gerald Edgar | Another small explanation may be needed: Since the hypothesis says it holds for nonnegative integrable $p$, you need to show it also holds for the delta distribution. | |
| S Oct 17, 2013 at 19:48 | history | suggested | Hans | CC BY-SA 3.0 |
Added a slightly more direct perspective for 1) in the same vein as svangen's original solution.
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| Oct 17, 2013 at 18:37 | vote | accept | Hans | ||
| Oct 17, 2013 at 18:34 | review | Suggested edits | |||
| S Oct 17, 2013 at 19:48 | |||||
| Oct 17, 2013 at 17:49 | history | edited | svangen | CC BY-SA 3.0 |
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| Oct 17, 2013 at 17:41 | comment | added | Hans | Also, a minor point, the sign of $k$ is determined by its integrability on the positive axis. | |
| Oct 17, 2013 at 17:40 | comment | added | Hans | Very nice, svangen. Thank you, for resolving largely question 1). The linearity of the functional $c$ you add gives the result for 1). That is mostly what I want. Do you have any thoughts on what happens if $c$ is nonlinear? Also, any thoughts on question 2)? | |
| Oct 17, 2013 at 17:33 | comment | added | svangen | The point I was trying to make was simply that for most choices of $c$, the equation would not have any solution $f$ (which wasn't clear to me apriori the way you had formulated the question). But I think we agree now. Let me know if you want me to elaborate on some step of the argument in this edit. | |
| Oct 17, 2013 at 17:29 | comment | added | Hans | Actually, your edited comment gives the proof for 1). | |
| Oct 17, 2013 at 17:28 | history | edited | svangen | CC BY-SA 3.0 |
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| Oct 17, 2013 at 17:24 | comment | added | Hans | The question says "where c[p] is a functional ...", so $c$ is A functional such that the first equation holds. The statement does not imply any functional $c$ would do. The statement does preclude the set of such $c$ being empty either. More directly, find $f$ such that there is some $c$ such that the first equation holds. Regarding your edited first comment, $c[\delta_z] = \exp(-kz)$ satisfies your condition $c[\delta_1]=c[\delta_{\frac{1}{2}}]$, or any similar condition. So it does not violate 1) and your conclusion does not hold. | |
| Oct 17, 2013 at 17:22 | history | edited | svangen | CC BY-SA 3.0 |
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| Oct 17, 2013 at 17:02 | comment | added | svangen | But maybe I'm misunderstanding your question (because I did not think of specifying $c$ as adding conditions, rather on the contrary). Do you intend to solve the equation for some given $c$, or for any $c$ (this was my initial interpretation), or find the $(c,f)$ that satisfies it? | |
| Oct 17, 2013 at 16:53 | history | edited | svangen | CC BY-SA 3.0 |
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| Oct 17, 2013 at 16:43 | comment | added | Hans | Even under your added condition of $c[p]=1, \forall p$, constant $f$ is still an exponential function with the exponent coefficient $k=0$, which is covered by the conclusion of 1), and not, violating it as you claimed. Besides, I am concerned about a less stringent condition of $c$ being a functional dependent on $p$. | |
| Oct 17, 2013 at 16:10 | history | answered | svangen | CC BY-SA 3.0 |