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Let $f$, $g$ be Riemann integrable functions on the interval $[a,b]$, that is $f,g \in \mathscr{R}([a,b])$.

(i) $\int_{a}^{b} (cf+g)^2\geq 0$ for all $c \in \mathbb{R}$.
(ii) $2|\int_{a}^{b}fg|\leq c \int_{a}^{b} f^2+\frac{1}{c}\int_{a}^{b} g^2$ for all $c \in \mathbb{R}^+$

I don't have a complete answer for either of these, but I have some ideas.

For (i) $\int_{a}^{b} (cf+g)^2=\int_{a}^{b}c^2f^2+2cfg+g^2=c^2\int_{a}^{b}f^2+c\int_{a}^{b}2fg+\int_{a}^{b}g^2$. The first third is positive because if $c <0$ then $c^2>0$. I'm not sure about the middle third. The final third is positive.

I did notice that if I can figure out (i)...(ii) follows from some rearranging.

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If your functions are real-valued, than the function $(cf+g)^2$ is nonegative (like every square is in $\mathbb{R}$). So is its integral (provided the bounds are in increasing order). – 1015 Feb 6 '13 at 1:08
I'm not sure how to argue this through rigorously. – emka Feb 6 '13 at 1:22
All you need to know is that $h\geq 0$ implies $\int h\geq 0$ (again, provided the bounds are in increasing order). Have you seen this fact? – 1015 Feb 6 '13 at 1:30
I have not seen that fact. – emka Feb 6 '13 at 1:49
I have seen you asked a question about Lebesgue dominated convergence theorem. It is hard to believe you got to that point without seeing the two ingredients needed here, namely $|\int h|\leq \int|h|$ and $\int h\geq 0$ when $h\geq 0$. I will see if I can find a reference. – 1015 Feb 6 '13 at 1:53

I will assume that the functions are real-valued. For the first point, I will use that if an R-integrable function $h$ is nonegative on $[a,b]$, then $\int_a^bh(x)dx\geq 0$. For the second point, I will use that if $h(x)\leq k(x)$ on $[a,b]$, then $\int_a^bh(x)dx\leq \int_a^bk(x)dx$. Note that the latter follows readily from the former by linearity of the integral.

1) We have $(cf(x)+g(x))^2\geq 0$ for all $x\in [a,b]$, so $\int_a^b(cf(x)+g(x))^2dx\geq 0$.

2) Recall that $2|ab|\leq a^2+b^2$ for every $a,b\in\mathbb{R}$. With $a=\sqrt{c}f(x)$ and $b=g(x)/\sqrt{c}$ ,this yields $$ 2|f(x)g(x)|\leq cf(x)^2+\frac{1}{c}g(x)^2 $$ on $[a,b]$. Hence $$ 2\int_a^b|f(x)g(x)|dx\leq c\int_a^bf(x)^2dx+\frac{1}{c}\int_a^bg(x)^2dx. $$ Finally, we have $|\int_a^bf(x)g(x)dx|\leq \int_a^b|f(x)g(x)|dx$, hence the second inequality.

Note: As you said, you can also deduce 2) from 1) directly by expanding $(cf+g)^2$ and then dividing by $c$. But $2|ab|\leq a^2+b^2$ is so useful I could not resist mentioning it.

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Why is it sufficient to show point 1 for just nonnegative Riemann integrable functions instead of for all real valued Riemann integrable functions? – emka Feb 7 '13 at 18:23
Point 1 is not true for a general real valued Riemann integrable function (try $h(x)=1$). – 1015 Feb 7 '13 at 19:03
I'm not sure I follow. For example $(cf+g)^2 \geq 0$ is always true, even if $f(x)=g(x)=1$ for all $x$. – emka Feb 7 '13 at 20:42
Yes. That`s not what I meant. Yes, $h=(cf+g)^2\geq 0$, which is why we can apply point 1 to deduce that $\int h\geq 0$. – 1015 Feb 7 '13 at 20:44
To clarify, something probably very stupid, $f$ and $g$ are any type of Riemann integrable functions (not necessarily nonnegative). $(cf+g)^2$ where $c$ is a real number and $f$ and $g$ are real-valued Riemann integrable functions is nonnegative and hence has nonnegative integral. – emka Feb 7 '13 at 21:29

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