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Aug 24, 2012 at 13:59 history edited Anand CC BY-SA 3.0
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Aug 24, 2012 at 13:58 comment added Anand Dear Robert Haslhofer, you are right. Thanks a lot. :-)
Aug 24, 2012 at 13:57 vote accept Anand
Aug 24, 2012 at 13:46 comment added Robert Haslhofer Since I am not sure if you really got the point, let me say it again more explicitly: If you compute the convolution of any reasonable function $f$ and a delta function, you simply get back the function itself! In formulas, $f * \delta_0=f$. (see Liviu's comment for when exactly the convolution is well defined).
Aug 24, 2012 at 13:24 comment added Anand Dear Robert Haslhoer, you are right. I need pairing of $f$ with $\delta_x$ for all $x\in R$, which equivalent to saying the convolution. :-)
Aug 24, 2012 at 12:32 comment added Robert Haslhofer Note that there is a difference between pairing $\langle f,\delta_0 \rangle = f(0)$ and convolution $f * \delta_0 = f$.
Aug 24, 2012 at 11:10 history edited Anand CC BY-SA 3.0
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Aug 24, 2012 at 10:31 answer added Liviu Nicolaescu timeline score: 5
Aug 24, 2012 at 8:15 history edited Anand CC BY-SA 3.0
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Aug 24, 2012 at 8:00 comment added Anand Dear Tapio Rajala, I am not looking for a norm on the vector space. After convolution, the function $x\mapsto (f*\delta_0)(x)$ might be viewed as a function in $L_loc^p(R)$, where $f$ is semicountinous.
Aug 24, 2012 at 7:23 comment added Tapio Rajala Certainly a pairing $\langle f, \mu\rangle = \int fd\mu$ is well defined for quite general vector spaces of functions and signed measures. It is still not clear to me what you are after. For example, do you want the vector spaces to have norms?
Aug 24, 2012 at 7:03 history edited Anand
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Aug 23, 2012 at 21:54 history edited Anand CC BY-SA 3.0
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Aug 23, 2012 at 21:42 comment added Anand Dear Professor Yemon Choi. You are right. In my mind, the paring for measures (e.g. $\delta_x$) on the one hand and continuous functions on the hand is legal. In my problem, I am thinking whether one can extend the above paring a bit from continuous functions to semicontinuous functions. I hope this makes question clear. Thanks a lot!
Aug 23, 2012 at 21:40 answer added timur timeline score: 2
Aug 23, 2012 at 21:32 comment added Yemon Choi You seem to be loose with what you mean by an inner product. What is the vector space on which you claim this inner product is well-defined? If you merely mean a pairing between two vector spaces, which ones do you have in mind?
Aug 23, 2012 at 21:31 history asked Anand CC BY-SA 3.0