Questions : [See below for context.]

$\rm\color{#c00}{a)}$ First, is the proof presented below $100$ % correct ?

$\rm\color{#c00}{b)}$ How would one justify the LHS of $(2)$ ? Are my thoughts correct ?

Thoughts : One way I see it is in three times : first we note that $(\cdot)^{1/p}$ is continuous, then we apply the Monotone Convergence Theorem (MON) and finally we note that $|\cdot|^p$ is continuous, so that all in all we can bring the limit all the way in (even if $\displaystyle\sum_{k\geq1} |f_k|$ is infinite in this case).

Another way I see it is by noting that every norm is (Lipschitz) continuous, so that we can effectively bring the limit inside (even if it is infinite in this case).

$\rm\color{#c00}{c)}$ Why must we take the absolute value of $f_k$ when investigating $\displaystyle\sum|f_k|$ ? Are my thoughts correct ?

Thoughts : I think that MON would not apply for $\rm\color{#c00}{b)}$ and I think that if we did not take the absolute value, then we could not speak, a priori, of the limit, that is, $\displaystyle\sum_{k\geq1} f_k$, in $(2)$. Taking the absolute value gives a series with positive terms and such a series either converges or diverges to $\infty$. A series with negative terms could diverge by oscillating.

$\rm\color{#c00}{d)}$ In $(4)$, defining $F$ by $0$ is purely arbitrary, right ? We could define $F$ to be anything on the null set, right ? Also, $\displaystyle\sum_{k\geq1} f_k$ does exist on the given set since absolute convergence implies convergence for series of numbers, right ?

$\rm\color{#c00}{e)}$ The equality $(5)$ is true since $$ F-\sum_{k=1}^{n}f_k=\sum_{k=n+1}^{\infty}f_k $$ almost everywhere, right ?

$\rm\color{#c00}{f)}$ Would it be fair to say that what was sufficient to prove was $\displaystyle\sum_{k\geq1}|f_k|<\infty$ a.e., [added] and that the reason the proof is not trivial is that $L^p(\mu)$ contains, by definition, finite-valued functions ? I mean, in measure theory we work with functions with range in the extended real line, $[-\infty,\infty]$, and if we allowed the functions in $L^p(\mu)$ to take infinite values, then in fact every series would be convergent (trivially) in $L^p(\mu)$. Is this correct ?

Def. : Let $(X,\cal{A},\mu)$ be a measured space. Then for $1\leq p<\infty$ we define $$ L^p(\mu):=\{f:X\to\mathbb{R}\text{ measurable and s.t. }\|f\|_p<\infty\}, $$ where $\|f\|_p:=\left(\int|f|^pd\mu\right)^{1/p}$. As usual we really take equivalence classes of functions differing only on a null set.

Thm (Riesz-Fischer) : $(L^p(\mu),\|\cdot\|_p)$ is complete for $1\leq p<\infty$.

Dem. : We know it suffices to show that every absolutely convergent series converges.

Let $(f_k)_{k\geq1}\subset L^p(\mu)$ be a sequence such that $$ \sum_{k=1}^{\infty}\|f_k\|_p<\infty.\tag{0} $$ Since, by Minkowski's inequality, one has $$ \left\|\sum_{k=1}^n|f_k|\right\|_p\leq\sum_{k=1}^n\|f_k\|_p,\tag{1} $$ letting $n\to\infty$ gives $$ \left\|\sum_{k=1}^{\infty}|f_k|\right\|_p\leq\sum_{k=1}^{\infty}\|f_k\|_p<\infty,\tag{2} $$ thus $$ \sum_{k=1}^{\infty}|f_k|<\infty\quad\text{a.e.}\tag{3} $$ Now define $$ F:=\begin{cases}\displaystyle\sum_{k\geq1}f_k&\text{on }\left\{\displaystyle\sum_{k\geq1}|f_k|<\infty\right\},\\0&\text{otherwise}.\end{cases}\tag{4} $$ Then $F$ is in $L^p(\mu)$ since $|F|\leq\displaystyle\sum_{k\geq1}|f_k|$ and by $(2)$. Also, $F$ is such that $$ \sum_{k=1}^{\infty}f_k=F, $$ because \begin{align} \left\|F-\sum_{k=1}^{n}f_k\right\|_p&=\left\|\sum_{k=n+1}^{\infty}f_k\right\|_p\tag{5}\\ &\leq\left\|\sum_{k=n+1}^{\infty}|f_k|\right\|_p\\ &\stackrel{(2)}{\leq}\sum_{k=n+1}^{\infty}\|f_k\|_p\xrightarrow[(n\to\infty)]{(0)}0.\quad\text{Q.E.D.} \end{align}


1 Answer 1

  1. Yes.
  2. This has nothing to do with continuity; only with the monotone convergence theorem. Note that we cannot use that continuity of the $L^p$-norm: In order to deduce $$\|g_n\|_p \to \|g\|_p$$ from the continuity of the norm, we already have to know that $g_n \to g$ in $L^p$ (recall that the norm is continuous as a mapping $\|\cdot\|: (L^p,\|\cdot\|_p) \to (\mathbb{R},|\cdot|)$. The argumentation goes as follows: By Minkowski's inequality, $$\left\| \sum_{k=1}^n |f_k| \right\|_p \leq \sum_{k=1}^n \|f_k\|_p \leq \sum_{k=1}^{\infty} \|f_k\|_p. \tag{1}$$ Since $|f_k| \geq 0$, we know $$\sum_{k=1}^n |f_k| \uparrow \sum_{k=1}^{\infty} |f_k| \qquad \text{as} \,\, n \to \infty. \tag{2}$$ Therefore, it follows from the monotone convergence theorem that $$\left\| \sum_{k=1}^{\infty} |f_k| \right\|_p = \sup_{n \in \mathbb{N}} \left\| \sum_{k=1}^n |f_k| \right\|_p \stackrel{(1)}{\leq} \sum_{k=1}^{\infty} \|f_k\|_p < \infty.$$
  3. Yes, that's correct. The series $\sum_{k=1}^{\infty} |f_k| \in [0,\infty]$ is well-defined since each addend is non-negative. Moreover, the non-negativity is crucial in $(2)$ (otherwise we cannot apply monotone convergence theorem).
  4. Yes.
  5. Yes.
  6. No, it's not that easy. If I understand you correctly, you would like to set $$F := \sum_{k=1}^{\infty} f_k$$ (as a pointwise limit) which might take values in $[-\infty,\infty]$. The problem is that the series $$\sum_{k=1}^{\infty} f_k$$ is not necessarily convergent, consider e.g. $f_k = (-1)^k$. This means that $F$ is not well-defined on some exceptional set and we have to ensure that this exceptional set is small (in $L^p$-sense). That's exactly what the proof above is about: We show that the exceptional set has measure $0$.
  • $\begingroup$ Concerning $\rm\color{#c00}{b)}$, I have some additional questions. $\rm\color{#c00}{ba)}$ We know (by definition) that $g_n\to g$ simply, but not that $g_n\to g$ in $L^p$ and that's why we can't use the continuity of the norm, right ? In fact strictly speaking $g\not\in L^p$ as $g$ might be infinite at some points, and we can't identify $g$ with a function in $L^p$ yet since that's what we are trying to prove, right ? So continuity of the norm doubly doesn't apply here. $\rm\color{#c00}{bb)}$ Is there any reason why you put $\sup$ instead of $\lim$ (even though it's equivalent here) ? $\endgroup$
    – Guest
    Jan 21, 2015 at 20:54
  • $\begingroup$ $\rm\color{#c00}{bc)}$ I don't understand why the continuity of $(\cdot)^{1/p}$ is not used before applying MON. MON tells you when you can interchange the limit with the integral, and the norm is the integral to the power $\frac{1}{p}$. In my question I said that we could bring the limit inside $(\cdot)^{1/p}$ even if the limit inside is infinite since sometimes we could not (e.g. $x^2-x$ is continuous but $\infty^2-\infty$ is undefined, or consider $\sin(x)$). $\endgroup$
    – Guest
    Jan 21, 2015 at 20:55
  • $\begingroup$ Finally, I thought what we really needed to verify in order to use MON is that $\left(\displaystyle\sum_{k=1}^n|f_k|\right)^p\uparrow\left(\displaystyle\sum_{k=1}^{\infty}|f_k|\right)^p$, which is indeed the case since $|\cdot|^p$ is increasing. Am I correct ? $\endgroup$
    – Guest
    Jan 21, 2015 at 20:55
  • $\begingroup$ @Guest ba) Yes, that's correct. bb) No, not really. Sometimes it's helpful (e.g. if you want to interchange limites), but in this case - as you said- in doesn't matter. bc) Yeah, you are right. And yes, we need that $|\cdot|^p$ is increasing in order to apply MON. $\endgroup$
    – saz
    Jan 22, 2015 at 5:35

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