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Let $$\large f_k(x)=\sum_{j=1}^{2^k}2^k\int_{I_{k,j}}f(t)\,dt\chi_{I_{k,j}}(x).$$

where $f\in L^p(0,1)$, and $I_{k,j}=\{x:(j-1)2^{-k}\leq x<j2^{-k}\}$, $j=1,\dots, 2^k$. In other words, $\{I_{k,j}\}$ are intervals of length $2^{-k}$ whose union is $[0,1)$.

My question is: How do we intuitively see that $\|f_k\|_p\leq\|f\|_p$? I can prove it rigorously (see below), but I currently lack the intuition to see/guess why it ought to be true.

$\|f_k\|_p\leq\|f\|_p$ turns out to be a key step in another question, and without this intuition, a priori I would not even think of proving this inequality. Hence, I am extremely curious of how to see/guess/suspect that $\|f_k\|_p\leq\|f\|_p$ intuitively

I can roughly see that $2^k\int_{I_{k,j}}f(t)\,dt$ is sort of an average of the values around the interval $I_{k,j}$.

Thanks for any help!

My proof:

\begin{align*} \int_0^1|f_k(x)|^p\,dx&=\int_0^1\left|\sum_{j=1}^{2^k}2^k\int_{I_{k,j}}f(t)\,dt\chi_{I_{k,j}}(x)\right|^p\,dx\\ &=2^{kp}\sum_{j=1}^{2^k}\int_{I_{k,j}}\left|\int_{I_{k,j}}f(t)\,dt\right|^p\,dx\\ &\leq 2^{kp}\sum_{j=1}^{2^k}\int_{I_{k,j}}\left[\int_{I_{k,j}}|f(t)|\,dt\right]^p\,dx\\ &\leq2^{kp}\sum_{j=1}^{2^k}\int_{I_{k,j}}\left[\left(\int_{I_{k,j}}|f(t)|^p\,dt\right)^{1/p}\left(\int_{I_{k,j}}|1|^{p'}\,dt\right)^{1/p'}\right]^p\,dx\tag{By Holder's inequality}\\ &=2^{kp}\sum_{j=1}^{2^k}\int_{I_{k,j}}\left[\left(\int_{I_{k,j}}|f(t)|^p\,dt\right)2^{-kp/p'}\right]\,dx\\ &=2^{kp-k-kp/p'}\sum_{j=1}^{2^k}\int_{I_{k,j}}|f(t)|^p\,dt\\ &=\int_0^1|f(t)|^p\,dt\\ &=\|f\|_p^p<\infty. \end{align*}

yoyostein
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    If you know conditional expectations, then D. Thomine's answer solves the problem http://math.stackexchange.com/questions/1990679/approximation-on-partitions-in-l20-1-times-omega – Davide Giraudo Nov 06 '16 at 09:59
  • Alas, I don't know measure theoretic probability. – yoyostein Nov 06 '16 at 11:36

1 Answers1

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This result doesn't have much to do with subintervals or sets having those specific measures. But as you suspected, it is important that we are dealing with averages of $f.$ That opens the door to convexity and Jensen's inequality.

Suppose $\mu$ is a positive finite measure on $X$ and $f\in L^1(X,\mu).$ Let $E_1, \dots , E_n$ be pairwise disjoint measurable subsets of $X$ of positive measure such that $X = \cup E_k.$ Let $g$ be a convex function on $\mathbb R.$ (In your setup, $g(t) = |t|^p,p\ge 1.$) Then

$$\int_X g\left (\sum_{k=1}^{n}(\frac {1}{\mu(E_k)}\int_{E_k} f\, d\mu) \cdot \chi_{E_k}\right)\,d\mu = \int_X \sum_{k=1}^{n}g\left (\frac {1}{\mu(E_k)}\int_{E_k} f\, d\mu \right) \cdot \chi_{E_k}\,d\mu$$

$$\le \int_X \sum_{k=1}^{n}\left (\frac {1}{\mu(E_k)}\int_{E_k} g\circ f\, d\mu \right) \cdot \chi_{E_k}\,d\mu = \sum_{k=1}^{n}\left (\frac {1}{\mu(E_k)}\int_{E_k} g\circ f\, d\mu \right) \cdot \mu (E_k) = \int_X g\circ f \, d\mu.$$

The appearance of the $\le$ sign above follows Jensen's inequality. This is exactly the desired inequality.

zhw.
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