Bourbaki

Exercise 7, § 4

Integration, VIII, p. 66

Let G be a locally compact group and β\beta a left Haar measure on G. Show that the algebra L1(G,β)L^1(G, \beta) has a unity element if and only if G is discrete. (Suppose G non-discrete, and let f0∈L1(G,β)f_0 \in L^1(G, \beta). There exists a compact neighborhood V of e such that

∫V∣f0(x)∣ dβ(x)<1. \int_V |f_0(x)|\, d\beta(x) < 1.
Let U be a symmetric compact neighborhood of e such that U2⊂VU^2 \subset V. Then, for almost every x∈Ux \in U,
∣(φU∗f0)(x)∣=∫U∣f0(y−1x)∣ dβ(y)≤∫V∣f0(x)∣ dβ(x)<1, |(\varphi_U * f_0)(x)| = \int_U |f_0(y^{-1}x)|\, d\beta(y) \leq \int_V |f_0(x)|\, d\beta(x) < 1,
thus f0f_0 is not a unity element for L1(G,β)L^1(G, \beta).)

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.