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Thread: Inequality of squares of integrable functions

  1. #1
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    Inequality of squares of integrable functions

    Hello,

    I am trying to show that $f,g \in \mathscr{R}$ implies that $(\int fg)^2\leq \int f^2 \int g^2$.

    The hint is to use a variant of the Cauchy-Schwartz inequality, $\sum a_i b_i \leq \sum a_i^2 \sum b_i^2$.

    We have a theorem which says that the Riemann integrability of $f$ and $g$ imply the integrability of the product fg so we can talk about $\overline{\int}fg$. This number is a lower bound for the upper sums $U(P,fg)$ for any partition $P$ of $[a,b]$ so, $\overline{\int}fg \leq U(P, fg)=\sum s_i t_i \Delta x_i$ (where $s_i$ is the max of $f$ on $[x_{i-1}, x_i]$, and $t_i$ is the max of $g$ on $[x_{i-1}, x_i]$.)

    So if I square both sides of the inequality I get $(\overline{\int}fg)^2 \leq (\sum s_i t_i \Delta x_i)^2$.

    It's pretty easy I think to argue that $\sum s_i t_i \Delta x_i < \sum s_i \Delta x_i t_i \Delta x_i$, since the $\Delta x_i$ are all positive. And I can write, $$(\overline{\int}fg)^2 \leq (\sum s_i t_i \Delta x_i)^2<(\sum s_i \Delta x_i t_i \Delta x_i)^2\leq (\sum s_i \Delta x_i)^2 (\sum t_i \Delta x_i)^2=U(P,f^2)U(P,g^2)$$

    Now I'm not to sure what to do. I have and expression which relates, eventually, the upper sums of a product $(\int fg)^2$ with the upper sums of $f^2$ and $g^2$.
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    Re: Inequality of squares of integrable functions

    Quote Originally Posted by bkbowser View Post
    Hello,
    I am trying to show that $f,g \in \mathscr{R}$ implies that $(\int fg)^2\leq \int f^2 \int g^2$.
    Because there are many, many approaches to the theory of integrals, I am not able to complete your proof. However. if we can assume that each of $f~\&~g$ are integrable on $[a,b]$ here is the proof I have used.

    We know that $f^2,~g^2,~\&~fg$ are also integrable. So let $A=\int {{f^2}}, B=2\int {{fg}},~\&~C=\int {{g^2}}$

    $ \begin{align*}q(x)&=Ax^2+Bx+C \\&=x^2\int {{f^2}}+2x\int {{fg}}+\int {{g^2}}\\&=\int {\left( {{x^2}{f^2} + 2xfg + {g^2}} \right)}\\&=\int {{{\left( {xf + g} \right)}^2}} \end{align*}$

    Now can you argue that $q(x)=\int {{{\left( {xf + g} \right)}^2}}\ge 0~?$
    If so that means that the quadratic $q(x)$ cannot be negative, so does not two real roots.
    Hence its discriminate is not positive:
    $ \begin{align*}B^2-4AC&\le 0 \\4{\left( {\int {fg} } \right)^2} - 4\left( {\int {{f^2}} } \right)\left( {\int {{g^2}} } \right)& \le 0\\{\left( {\int {fg} } \right)^2}&\le \left( {\int {{f^2}} } \right)\left( {\int {{g^2}} } \right) \end{align*}$
    Thanks from SlipEternal
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    Re: Inequality of squares of integrable functions

    That's pretty nice.

    Thanks
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