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Exercise 1.3.19 (Lebesgue integration is linear)

Let L1 (d ) be the (vector) space of absolutely integrable functions. Show that the integration : L1 is a *-linear transformation.

Answers

Let f,g L1 (d ) be absolutely integrable Lebesgue measurable functions, and let c . We demonstrate the properties of an -linear transformation. The steps of this proof will strongly resemble the analogous proof of Exercise 1.3.2.

(i)
(additivity)
First, suppose that f is real-valued. By definition we have d(f + g) =d(f + g)+ d(f + g) Recall that in Exercise 1.3.2 we have verified the identity  (f + g)+ (f + g) = (f+ f) + (g+ g). Rewriting it as  (f + g)+ + f + g = (f + g) + f+ + g+ allows us to conclude that  (f + g)+ +f +g =(f + g) +f+ +g+ by Corollary  1.3.14. =d(f + g) +df+ +dg+ dfdgd(f + g) = [df+ df] + [dg+ dg] =df +dg

Now suppose that f is complex valued. We then have

d(f + g) =d(f + g) + id(f + g) =d(ℜf + ℜg) + id(ℑf + ℑg) =dℜf +dℜg + idℑf + idℑg = [dℜf + idℑf ] + [dℜg + idℑg] =df +dg
(ii)
(homogeneity)
As usual, first suppose that f is real valued, and c . We then have dcf =d(cf)+ d(cf) We have two cases: either c 0; in which case we one can verify case-wise the  identity (cf)+ = c(f)+ and (cf) = c(f); and by Exercise 1.3.10 (iii) we  obtain =dc(f)+ dc(f) = c [df+ df] = cdf Or we have c < 0 in which case it is verifyable that (cf)+ = c(f) and  (cf) = c(f)+ =d c(f)d c(f)+ = c [dfdf+] = cdf

Now suppose that f is complex valued. The scalar multiplication property is demonstrated as follows. We can rewrite cf as

cf(x) = [(c)+iℑ(c)][(f(x))+iℑ(f(x))] = (c)(f(x))(c)(f)+iℑ(c)(f)+iℜ(c)(f)

Using this identity we obtain

dcf =d(cf) + i d(cf) =d [(c)(f) (c)(f) ] + i d [(c)(f) + (c)(f) ] = (c)d(f) (c)d(f) + iℑ(c) d(f) + (c)d(f) =d(f) [(c) + iℑ(c)] + id(f) [(c) 1iℑ(c)] = [(c) + iℑ(c)] (d(f) + id(f)) = c df
(iii)
(conjuctivity)
The conjuctive property follows as follows df¯ =d(f¯) + i d(f¯) =d(f) + i d(f) = d(f) + i d(f)¯ = df¯
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2020-08-30 00:00
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