For every integrable $f: (X,\mu) \to \mathbb{R}^n$, $||\int f d\mu|| \le \int ||f||d\mu$.

Question

For every integrable $f: (X,\mu) \to \mathbb{R}^n$, $||\int f d\mu|| \le \int ||f||d\mu$, where $||\cdot ||$ is the standard Euclidean norm.

Expanding the first one, I get $\sqrt{\int f_1^2 + \cdots + \int f_n^2}$, but I don't know which inequalities I need to use to bound this on the right value. How can I show this? I would greatly appreciate any help.

score 0 · Answer 1 · 2021-09-03T23:18:52.480

0

If $\mu(X)<\infty$, then this follows from Jensen's inequality ($\because x\mapsto\lVert x\rVert$ is convex), i.e., $$ [\mu(X)]^{-1}\left\lVert\int f d\mu\right\rVert=\left\lVert [\mu(X)]^{-1}\int f d\mu\right\rVert \le [\mu(X)]^{-1}\int \lVert f\rVert d\mu. $$

edited Sep 03 '21 at 23:18

answered Apr 28 '16 at 05:12

As far as I know Jensen's inequality is valid for random variables taking values in an open interval. How can we modify it to fit this case? – nomadicmathematician Feb 03 '17 at 15:08
What is the multivariate Jensen's inequality exactly? – nomadicmathematician Feb 04 '17 at 02:39

For every integrable $f: (X,\mu) \to \mathbb{R}^n$, $||\int f d\mu|| \le \int ||f||d\mu$.

1 Answers1