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.

pairing$\langle f,\delta_0 \rangle = f(0)$ andconvolution$f * \delta_0 = f$. $\endgroup$3more comments