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Dear all,

It is clear that if $f:R\mapsto R$ is a continuous function, than $< f, \delta_x >=f(x)$. Now, if $f$ is only semicontinous, can we say that $< f, \delta_x >=f(x)$? I think this is true at all continuous points of $f$. But when $f$ has a jump at $x$, can we properly define this inner product? Does anyone know any references dealing with this matter?

By the way, I checked the wikipedia page about semicontinuous functions, from where I find Bourbaki's two volumns. But I didn't find any information about such pairing.

EDIT: Following the remark by Tapio Rajala, I think what I want is the following:

Suppose $f$ is a semicontinous function. Then the function

$$ x\mapsto (f* \delta_0 )(x) = \int f(x-y) \delta_0(y)d y $$

is in $L^{\infty}_{loc}(R)$.

It seems true for me. If anyone knows a reference, it would be nice, even though the proof seems not difficult. :-)

Here is another motivation of this problem:

Consider the wave equation in $R$

$$ \frac{\partial^2 }{\partial t^2} u(t,x) = \frac{\partial^2 }{\partial x^2} u(t,x) ,\quad t>0,x\in R\;, $$

with vanishing initial position. Suppose that the initial velocity is a nonnegative Borel measure $\mu$. The solution is

$$ u(t,x) = \mu\left([x-t,x+t]\right) = (\mu * G_t)(x)\:, $$

where $G_k(x)$ is the fundamental solution:

$$ G_t(x) = 1_{|x|\le t} $$

which is a semicountinuous function. In particular, by letting $\mu=\delta_0$, we have the problem of pairing a delta function with a semicontinuous function. What I need is simply that the solution $u(t,x)$ is in $L^\infty_{loc}(R)$. I think this is true.

Thanks a lot for any hints and helps!

RIP, Bill.

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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? – Yemon Choi Aug 23 '12 at 21:32
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! – Anand Aug 23 '12 at 21:42
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? – Tapio Rajala Aug 24 '12 at 7:23
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. – Anand Aug 24 '12 at 8:00
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). – Robert Haslhofer Aug 24 '12 at 13:46
up vote 5 down vote accepted

This is an ill posed problem. The Dirac $\delta$ is a continuous linear map from the (locally convex) space of continuous function on $\mathbb{R}^n$ to $\mathbb{R}$. You ask if it admits an extension to the larger set of semicontinuous functions. First, semicontinuity is not a linear condition: the sum of a lower semicontinuous function with an upper semi-continuous function may not be semi-continuous in any fashion. (Lower semicontinuous functions do form a convex cone.) I assume you want your extension to be linear. So you may be asking for an extension of $\delta$ as a linear function on the larger vector space of functions which can be represented as a sum of a lower semi-continous function with an upper semi-continous function. Extensions of linear functionals from a subspace do exist and by no means are unique. Even requiring some sort of continuity (?) of the extension, the result will not be unique. Without specifying what properties you expect your extension to have you cannot expect a definitive answer. You need to phrase your question more accurately.

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Dear Professor Liviu Nicolaescu, thank you very much for your answer. I will add another motivation in my post. My motivation is not so abstract as what you are thinking. Thanks a lot! :-) – Anand Aug 24 '12 at 11:02
Anad, now I understand your motivation. The convolution of two distributions is well defined provided that their supports satisfy a certain, easy to verify condition. (This condition is satisfied in your case.) A good place to learn about this is Chap. 11 of the new book Distributions. Theory and Applications by J.J. Duistermaat and J.A.A.C Kolk, Birkauser, 2010, written for advanced undergraduates. Sec. 13.2 of the same book discusses in great detail the wave equation. If you can read French, then have a look at L. Schwartz' Theorie des Distributions. It's a masterpiece. – Liviu Nicolaescu Aug 24 '12 at 12:44
Dear Professor Liviu Nicolaescu, thanks a lot for your references. I am now clear. :-) – Anand Aug 24 '12 at 14:06

No, but you can define various regularizations, such as the left limit, right limit, and their averages.

Update: Now that you gave concrete motivations, it is clearer what you were after. Indeed, the convolution $\delta*f$ makes sense for any distribution $f$, and equal to $\delta*f=f$. So no problem with your examples.

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Dear Timur, thanks for your answer. I have the same thinks as you. But do you know some references that I don't need to invent or care about everything? Thanks a lot! – Anand Aug 23 '12 at 21:45
A semi-continuous function needs to have neither left nor right limits (though it is continuous on a dense $\mathcal{G}_\delta$-set). – Wolfgang Loehr Aug 24 '12 at 9:49
Dear Wolfgang Loehr, could you please clarify a bit your comments? So you don't agree what timur said? Thanks a lot. – Anand Aug 24 '12 at 11:14
Consider $f(x):=\sin(\frac1x)$ for $x\ne0$, $f(0):=-1$. Then clearly $f$ is lower semi-continuous but has neither a left nor a right limit at $0$. – Wolfgang Loehr Aug 24 '12 at 14:28
@Wolfgang: Thanks for your comment. So left and right limits exist for a smaller class of functions, e.g. for piecewise continuous functions. In any case, now the OP gave some concrete motivations, it appears that convolution with $\delta$ is what he is after, not $\delta$ itself. – timur Aug 24 '12 at 14:44

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