Suppose the domain of the propositional function Px consists

Suppose the domain of the propositional function P(x) consists of -5, -3, -1, 1, 3 and 5. Express these statements without using quantifiers, instead using only negations, disjunctions and conjunctions. (a) xP(x) (b) Forall x P(x) (c) Forall x((x notequato 1) rightow P(x)) (d) x((x greaterthanorequalto 0)^P(x)) (e) x(not P(x))^Forall x((x

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Using this thread, and also the indisputable fact that if f1f1 and f2f2 ar 2 integrable functions, F(fg)=F(f)F(g)F(fg)=F(f)F(g), we have
F(ddx(fg))(x)=ixF((fg))(x)=ixF(f)(x)F(g)(x),
F(ddx(fg))(x)=ixF((fg))(x)=ixF(f)(x)F(g)(x),
and
F((ddxf)g)(x)=(F(ddxf))(F(g)(x))=ixF(f)(x)F(g)(x).
F((ddxf)g)(x)=(F(ddxf))(F(g)(x))=ixF(f)(x)F(g)(x).
We conclude by individuality of Fourier remodel.

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answered Jul thirty one \'12 at 16:57

Davide Giraudo
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How you\'ll be able to take Fourier remodel whereas we do not comprehend it has Fourier remodel or not? In alternative words, we do not understand ddx(fg)ddx(fg) is in L1L1? For the second Fourier remodel, it\'s correct since we all know that fgfg is in L1L1. – rfvahid Jul thirty one \'12 at 17:09
     
Indeed, it deserves additional details. i feel associate degree approximation argument will work (approximate in L1L1 ff and gg by C1C1 functions with compact support). – Davide Giraudo Jul thirty one \'12 at 17:20
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Note that, if fL1(R)fL1(R) then it\'s Fourier translatable. Since,

f(x)eixw|f(x)|<
|f(x)eixw||f(x)|<
.

To prove that the convolution of 2 L1(R)L1(R) performs is once more associate degree L1(R)L1(R) function, let

h(x)=f(t)g(xt)dt
h(x)=f(t)g(xt)dt
|h(x)|dx|f(t)||g(xt)|dtdx=|f(t)||g(xt)|dxdt=|f(t)|||g||1dt=||f||1||g||1hL1(R).
|h(x)|dx|f(t)||g(xt)|dtdx=|f(t)||g(xt)|dxdt=|f(t)|||g||1dt=||f||1||g||1hL1(R).
The modification of the order of integration is even by Fubini\'s theorem. So, you\'ll be able to use the Fourier technique as in Davide\'s answer.

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edited August one \'12 at 16:30
answered Jul thirty one \'12 at 18:46

Mhenni Benghorbal
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Definition:
h(x)=fg(x)=f(xt)g(t)dt
h(x)=fg(x)=f(xt)g(t)dt
Let\'s calculate derivative:

dhdx=limdx>0(f(x+dxt)g(t)dtf(xt)g(t)dt)dx=limdx>0((f(x+dxt)f(xt))dxg(t)dt
dhdx=limdx>0(f(x+dxt)g(t)dtf(xt)g(t)dt)dx=limdx>0((f(x+dxt)f(xt))dxg(t)dt
If assume that for
(f(x+dxt)f(xt))dx
(f(x+dxt)f(xt))dx
exist some integrable perform q(t), that is edge on this expression, perhaps except potential on set with live zero then by Lebesgue dominated convergence theorem we will push the limit within integral.

dhdx=fg(x)=f(xt)g(t)dt=fg