Answer to Question #133626 in Real Analysis for PRATHIBHA ROSE C S

Question

Prove that L^p (mu)is complete both in descrete and infinite case.

Answer 1

Let’s prove completeness for the classical Banach spaces, say L^p[0,1] where p≥1.

Since the case p=∞ is elementary, we may assume 1≤p<∞. Let [f⋅]∈(L^p)^𝐍

 be a Cauchy sequence. Define [g₀]:=[f₀] and for n>0 define [g_n]:=[f_n-f_n-1]. Then [∑^N_n=₀g_n]=[f_N]

 and we see that

∞∑_n=0∥g_n∥=∞∑_n=0∥f_n-f_n-1∥≤???<∞.

Thus it suffices to prove that etc.

It suffices to prove that each absolutely summable series in L^p is summable in L_p  to some element in L^p

Let {f_n} be a sequence in L^p with ∑_n=1^∞∥f_n∥=M<∞, and define functions g_n by setting g_n⁢(x)=∑_k=1ⁿ|f_k⁢(x)|. From the Minkowski inequality we have

∥g_n∥≤∑_k=1ⁿ∥f_k∥≤M.

Hence

∫g_n^p≤M^p.

For each x, {g_n⁢(x)} is an increasing sequence of (extended) real numbers and so must converge to an extended real number g⁢(x). The function g so defined is measurable, and, since g_n≥0, we have

∫g^p≤M^p

by Fatou’s Lemma. Hence g^p is integrable, and g⁢(x) is finite for almost all x.

For each x such that g⁢(x) is finite the series ∑_k=1^∞f_k⁢(x) is an absolutely summable series of real numbers and so must be summable to a real number s⁢(x). If we set s⁢(x)=0 for those x where g⁢(x)=∞, we have defined a function s which is the limit almost everywhere of the partial sums s_n=∑_k=1ⁿf_k. Hence s is measurable. Since |s_n⁢(x)|≤g⁢(x), we have |s⁢(x)|≤g⁢(x). Consequently, s is in L^p and we have

|s_n⁢(x)-s⁢(x)|^p≤2^p⁢[g⁢(x)]^p.

Since 2^p⁢g^p is integrable and |s_n⁢(x)-s⁢(x)|^p converges to 0 for almost all x, we have

∫|s_n-s|^p→0

by the Lebesgue Convergence Theorem. Thus ∥s_n-s∥^p→0, whence ∥s_n-s∥→0. Consequently, the series {f_n} has in L^p the sum s.