I m working on the proof the following corollary for ages... I would appreciate any help!!
Cor: Let $h : [0, \infty ) \rightarrow \mathbb{R}$. We define the convolution operator $*$ for the functions $h*m$ and $h*F$ by: $$ (h*m)(t):=\int\limits_{0}^{t} h(t-x)dm(x), \quad (h*F)(t):=\int\limits_{0}^{t} h(t-x)dF(x),$$ whenever these integral exists. Then we have: $(h*m)*F=h*(m*F).$
I know the definition of Riemann-Stjeltes integral. I also understand the proof of the associativity of convolution operator. But here are differential form, where i have so many problems by calculating.
I have shown the commutativity, so i can use it in the proof. This is what i did and where i stuck:
LHS:
$[h*(m*F)](t)=\int\limits_{0}^{t} h(t-x)d(m*F)(x) =-\int\limits_{0}^{t} (m*F)(t-x)dh(x) =-\int\limits_{0}^{t} \left[ \int\limits_{0}^{t-x} m(t-x-y)dF(y) \right] dh(x) =\int\limits_{0}^{t} \left[ \int\limits_{0}^{t-x} F(t-x-y)dm(y) \right] dh(x) =\int\limits_{0}^{t} \int\limits_{0}^{t-x} F(t-(x+y)) dm(y) dh(x) =\int\limits_{0}^{t} \int\limits_{x}^{t} F(t-u) dm(u-x) dh(x) =\int\limits_{0}^{t} \int\limits_{0}^{u} F(t-u) dh(x) dm(u-x) $
RHS:
$[(h*m)*F](t) =\int\limits_{0}^{t} (h*m)(t-u)dF(u) =\int\limits_{0}^{t} \left[ \int\limits_{0}^{t-u} h(t-u-y)dm(y) \right] dF(u) =\int\limits_{0}^{t} \left[ \int\limits_{u}^{t} h(t-u-y)dm(y) \right] dF(u) = \int\limits_{0}^{t} \int\limits_{u}^{t} h(t-u-y)dm(y) dF(u) = \int\limits_{0}^{t} \int\limits_{0}^{u+y} h(t-u-y) dF(u) dm(y) = - \int\limits_{0}^{t} \int\limits_{0}^{u+y} F(t-(u+y)) dh(u) dm(y) $
So i don't get the same, where is my mistake? please i need any help!
Thank you very much for any help!
Using the commutativity and the definition of the convolution operator, we find
$$\begin{align*} h \ast (m \ast F)(t) &= (m \ast F) \ast h (t) \\ &= \int_0^t (m \ast F)(t-x) \, dh(x) \\ &= \int_0^t \int_0^{t-x} m((t-x)-y) \, dF(y) \, dh(x) \\ &= \int_0^{\infty} \int_0^{\infty} 1_{[0,t]}(x) \cdot 1_{[0,t-x]}(y) \cdot m(t-x-y) \, dF(y) \, dh(x). \end{align*}$$
From
$$1_{[0,t]}(x) \cdot 1_{[0,t-x]}(y) = 1_{[0,t]}(x)\cdot 1_{[0,t-x]}(y)\cdot 1_{[0,t]}(y) = 1_{[0,t-y]}(x)\cdot 1_{[0,t]}(y)$$
we therefore conclude by Tonelli's theorem
$$\begin{align*} h \ast (m \ast F)(t) &= \int_0^{\infty} \int_0^{\infty} 1_{[0,t-y]}(x)\cdot 1_{[0,t]}(y)\cdot m(t-x-y) \, dh(x) \, dF(y) \\ &= \int_0^t \int_0^{t-y} m((t-y)-x) \, dh(x) \, dF(y) \\ &= \int_0^t (m \ast h)(t-y) \, dF(y) \\ &= (m \ast h) \ast F(t). \end{align*}$$
Using again the commutativity, this finishes the proof.