I am trying to work through a homework set, and the one of the problems states the following inequity:
$|\int \prod_{j=1}^{2^n} f_{j}d\mu| \leq \prod_{j=1}^{2^n} (\int |f_{j}|{ ^2}^n d\mu)^{1/{2^n}}$.
How can I prove it? Applying Cauchy-Schwarz?
Thank you for any help
On
This is a special case of Generalized Holder's inequality. See, for example, https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwiq0qfE2_XeAhWLMo8KHe-zDNMQFjABegQIBBAC&url=https%3A%2F%2Fwww.emis.de%2Fjournals%2FHOA%2FIJMMS%2F26%2F17.pdf&usg=AOvVaw3GgOogQcE7bhWFp4Rx0B4g
Firstly, note that
$-|f(x)| \leq f(x) \leq |f(x)| \implies -\int|f(x)| \leq \int f(x) \leq \int|f(x)| \implies |\int f(x)| \leq \int|f(x)|$
Then,
$|\int \prod_{j=1}^{2^n} f_{j}d\mu| \leq \int \prod_{j=1}^{2^n} |f_{j}|d\mu$
Applying the Cauchy Schwarz Inequality $\Big(\int f(x) g(x) \leq \sqrt{\int f^{2}(x)}\sqrt{\int g^{2}(x)}\Big)$
$\int \prod_{j=1}^{2^n} |f_{j}|d\mu \leq \big(\int\prod_{j=1}^{2^{n-1}} |f_{j}|^2d\mu\big)^{\frac{1}{2}} \big(\int \prod_{j=2^{n-1}+1}^{2^n} |f_{j}|^2d\mu\big)^{\frac{1}{2}}$
Applying it to each of the individual integrals on the RHS,
$\big(\int\prod_{j=1}^{2^{n-1}} |f_{j}|^2d\mu\big)^{\frac{1}{2}} \big(\int \prod_{j=2^{n-1}+1}^{2^n} |f_{j}|^2d\mu\big)^{\frac{1}{2}} \leq$ $ \big(\int \prod_{j=1}^{2^{n-2}} |f_{j}|^{2^2}d\mu\big)^{\frac{1}{4}} \big(\int \prod_{j=2^{n-2}+1}^{2^{n-1}} |f_{j}|^{2^2} \big)^{\frac{1}{4}} \big(\int \prod_{j=2^{n-1}+1}^{2^{n}-2^{n-2}+1} |f_{j}|^{2^2}d\mu\big)^{\frac{1}{4}} \big(\int \prod_{j=2^{n}-2^{n-2}+2}^{2^{n}} |f_{j}|^{2^2}d\mu\big)^{\frac{1}{4}} $
Now because there are $2^n f_j$, we can keep "halving" the individual integrals until we reach integral of one ($2^0$) $f_j$ with Cauchy Schwarz as I did above. Iterated application should end up with
$|\int \prod_{j=1}^{2^n} f_{j}d\mu| \leq \big(\int |f_{1}|^{2^n}d\mu\big)^{\frac{1}{2^n}} \times \dots \times \big(\int |f_{2^n}|^{2^n}d\mu\big)^{\frac{1}{2^n}} = \prod_{j=1}^{2^n} (\int |f_{j}|{ ^2}^n d\mu)^{1/{2^n}} \tag*{$\blacksquare$}$