Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

1 Answer 1

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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.