Can we calculate the inner product of a semicontinous function with the Dirac delta function? 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.
 A: 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.
A: 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.
