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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!

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  • $\begingroup$ I doubt that the definition of convolution in the description. Why are $m(x)$ and $F(x)$ behind the differential operator? $\endgroup$
    – Magician
    Jun 9 '14 at 8:02
  • $\begingroup$ It is right. $ (f*g)(t)=\int\limits_{0}^{t} f(t-x)dg(x)=\int\limits_{0}^{t} f(t-x)g'(x)dx$. $\endgroup$
    – mr_T
    Jun 9 '14 at 8:06
  • $\begingroup$ is this your own definition? $\endgroup$
    – Magician
    Jun 9 '14 at 14:13
  • $\begingroup$ no... it's from a book $\endgroup$
    – mr_T
    Jun 10 '14 at 8:01
  • $\begingroup$ please give me the reference. It contradicts my experience. $\endgroup$
    – Magician
    Jun 10 '14 at 19:44
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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.

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  • $\begingroup$ Thank you very much saz!!!! I understand it! $\endgroup$
    – mr_T
    Jun 10 '14 at 8:52
  • $\begingroup$ sorry I do have a question. $1_{[ 0,t ]}(x) 1_{[ 0,t-x ]}(y)=1_{[ 0,t ]}(x) 1_{[ 0,t-x ]}(y) 1_{[ 0,t ]}(y).$ is clear to me. but why is $1_{[ 0,t ]}(x) 1_{[ 0,t-x ]}(y) 1_{[ 0,t ]}(y)=1_{[ 0,t-y ]}(x) 1_{[ 0,t ]}(y)$ ? $\endgroup$
    – mr_T
    Jun 10 '14 at 9:30
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    $\begingroup$ Well, because $$y \leq t-x \Leftrightarrow x \leq t-y$$ i.e. $1_{[0,t-x]}(y) = 1_{[0,t-y]}(x)$ $\endgroup$
    – saz
    Jun 10 '14 at 11:03
  • $\begingroup$ thanx a lot!!!! $\endgroup$
    – mr_T
    Jun 10 '14 at 11:05
  • $\begingroup$ @mr_T You are welcome. $\endgroup$
    – saz
    Jun 10 '14 at 11:23

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