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I’m working on Nualart’s book “The Malliavin calculus and related topics” and in the proof of lemma 1.1.3 he mentions that the operators $P_n$ have their operator norm bounded by 1. I fail to see why, can you help me? Using Jensen’s inequality I get a norm more akin to $2^n$, so I guess Jensen is too weak to prove that?

Quoting the proof:

Let $u$ be a process in $L^2_a([0,1]\times\Omega)$ ($L^2_a$ are the adapted processes w.r.t Brownian motion) and consider the sequence of processes defined by $\tilde u^n(t)=\sum_{i=1}^{2^n-1}2^n\left(\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)ds\right)1_{]i2^{-n},(i+1)2^{-n}]}(t)$.

We claim that the sequence converges to $u$ in $L^2([0,1]\times\Omega)$. In fact define $P_n(u)=\tilde u^n$. Then $P_n$ is a linear operator in $L^2([0,1]\times\Omega)$ with norm bounded by one.

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2 Answers 2

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I believe that there are some typos. It should be \begin{align*} \tilde u^n(t)=\sum_{i=1}^{\color{red}{2^n}}2^n\left(\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)ds\right)1_{]\color{red}{(i-1)2^{-n}, i2^{-n}}]}(t) \end{align*} Then, note that \begin{align*} E\left(\int_0^1\left(\tilde{u}^n\right)^2dt \right)&=\sum_{i=1}^{2^n}2^{2n}E\left(\int_{(i-1)2^{-n}}^{i2^{-n}}\left(\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)ds\right)^2 dt\right)\\ &=\sum_{i=1}^{2^n}2^{n}E\left(\left(\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)ds\right)^2 \right)\\ &\le \sum_{i=1}^{2^n}2^{n}E\left(\int_{(i-1)2^{-n}}^{i2^{-n}}1^2ds\int_{(i-1)2^{-n}}^{i2^{-n}}u^2(s)ds \right) \ \ \text{(by Schwarz inequality)}\\ &= \sum_{i=1}^{2^n}E\left(\int_{(i-1)2^{-n}}^{i2^{-n}}u^2(s)ds\right)\\ &= E\left(\sum_{i=1}^{2^n}\int_{(i-1)2^{-n}}^{i2^{-n}}u^2(s)ds\right)\\ &= E\left(\int_0^{1}u^2(s)ds \right)\\ &=||u||^2. \end{align*} Therefore, $P$ has norm 1.

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  • $\begingroup$ Can you explain the last inequality? How does the factor of $2^n$ vanish? $\endgroup$
    – dlrlc
    Commented Nov 1, 2016 at 8:12
  • $\begingroup$ I added more details. $\endgroup$
    – Gordon
    Commented Nov 1, 2016 at 12:43
  • $\begingroup$ So Cauchy-Schwarz is the inequality needed, not Jensen… now I feel stupid. Thank you very much! $\endgroup$
    – dlrlc
    Commented Nov 1, 2016 at 12:46
  • $\begingroup$ @Franzo: Many thanks for the bounty. $\endgroup$
    – Gordon
    Commented Nov 1, 2016 at 20:46
  • $\begingroup$ Enjoy! Thanks for the answer! $\endgroup$
    – dlrlc
    Commented Nov 1, 2016 at 20:56
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Jensen inequality works fine. Indeed (correcting the mistake spotted by Gordon):

$$\int_0^1 (\tilde{u}^n)^2 (t) dt = \int_0^1 \sum_{i=0}^{2^n-1}\left(\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)2^nds\right)^2 1_{]i2^{-n},(i+1)2^{-n}]}(t) dt,$$

since all the caracteristic functions have disjoint support. Hence, using Jensen inequality (with the probability measures $2^n ds$ on each interval $]i2^{-n},(i+1)2^{-n}]$):

$$\begin{align}\|P_n u\|_{\mathbb{L}^2}^2 & \leq \int_0^1 \sum_{i=0}^{2^n-1}\left(\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)^2 2^nds\right) 1_{]i2^{-n},(i+1)2^{-n}]}(t) dt \\ & = \sum_{i=0}^{2^n-1} \left(2^n\int_{(i-1)2^{-n}}^{i2^{-n}}u(s)^2 ds\right) \int_0^1 1_{]i2^{-n},(i+1)2^{-n}]}(t) dt \\ & = \sum_{i=0}^{2^n-1} \int_{(i-1)2^{-n}}^{i2^{-n}}u(s)^2 ds \\ & = \|u\|_{\mathbb{L}^2}^2. \end{align}$$

So, in this case, Jensen's inequality is not weaker; you just need to be careful on where you apply it.

Note, by the way, that this can be proved much faster. Fix $n\geq 0$. Let $\pi_n := \{]i2^{-n},(i+1)2^{-n}]: \ 0 \leq i < 2^n\}$, and $\mathcal{C}_n := \sigma (\pi_n)$ be the $\sigma$-algebra generated by $\pi_n$. Then:

$$P_n (u) = \mathbb{E} (u | \mathcal{C}_n),$$

and the conditional expectation is always a weak $\mathbb{L}^2$ contraction, which can be proved for instance with the conditional version of Jensen inequality:

$$\mathbb{E} (P_n (u)^2) = \mathbb{E} (\mathbb{E} (u | \mathcal{C}_n)^2) \leq \mathbb{E} (\mathbb{E} (u^2 | \mathcal{C}_n)) = \mathbb{E} (u^2).$$

This point of view makes more sense from a probabilist's standpoint, I think.

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  • $\begingroup$ Clever usage of Jensen here. Hope you don’t mind me giving the bounty to the first answer. $\endgroup$
    – dlrlc
    Commented Nov 1, 2016 at 17:48

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