1
$\begingroup$

Show that if $f,g: [a,b] \rightarrow \mathbb{R}$ are Riemann integrable then $\sqrt{f^2+g^2}$ is Riemann integrable and

$\int_a^b \sqrt{f^2 + g^2} dx \leq \int_a^b |f| dx + \int_a^b |g| dx$

So I'll start by writing down what I know:

I know by arithmetic of integrals that $|\int_a^b fg dx| \leq \sqrt{\int_a^b f^2} dx \cdot \sqrt{\int_a^b g^2 dx}$.

I also know that the function will be RI iff: $\exists P=\{a=x_0, x_1, ... , x_n = b\} \text{ a partition}$ s.t $U(f,p) - L(f,p) < \epsilon$. Finally, if f is any bounded function then $\overline{\int} f dx \geq \underline{\int} f dx$.

And so to prove Riemann integrability I'm guessing we let choose partitions such that:

$U(f,p_1) - (\text{ some } \epsilon ) < \int_a^b fdx < L(f,p_1) + (\text{ some } \epsilon)$. Similarly for $p_2$.

Let $P = P_1 \cup P_2$ band let $x,y \in [X_{i-1}, x_i]$ for some $1 \leq i \leq N$. And so we're left with

$\sqrt{(f^2(x) + g^2(x)) - ((f^2(y)-g^2(y))}$

I know I'm supposed to show that the above is less than something, and do an $\epsilon$ argument, but I really can't figure it out.

As for proving the inequality,

$\sqrt{U(f^2+g^2,p)} \leq \sqrt{\sum \sup f^2+g^2 \Delta x_i}$ And again, I don't know how to proceed.

Help would be much appreciated!

$\endgroup$
3
$\begingroup$

$f,g$ are Riemann-Integrable $\implies f^2,g^2$ are Riemann-Integrable

$f,g$ are Riemann-Integrable $\implies f+g $ are Riemann-Integrable

$ h$ is Riemann-Integrable $\text {and} h\ge 0 \implies \sqrt h$ is Riemann-Integrable

Now combine all three.

$\endgroup$
  • $\begingroup$ I think this is the simplest way to go. $\endgroup$ – zhw. Dec 8 '16 at 5:28
1
$\begingroup$

We have $\sqrt{a^2+b^2}\leq |a|+|b|$. See this simply by squaring each side. Now, monotonicity of the integral implies your result.

$\endgroup$
  • $\begingroup$ I'm probably being really stupid, but I'm still stuck. So we have $\sqrt{(f^2(x) + g^2(x)) - ((f^2(y)-g^2(y))} \leq |f(x)+g(x)| + |-f(y) + g(y)|$. I'm guessing I want to set a M = sup (something with f(x), g(x)) so we can further bound the above equation. $\endgroup$ – Nikitau Dec 8 '16 at 4:15
  • $\begingroup$ If you have Rudin's PMA, I'm suggesting you apply Theorem 6.12 (b) $\endgroup$ – Half-Pint Dec 8 '16 at 17:25
  • $\begingroup$ Yup, I do have Rudin's PMA. I was trying to mimic a proof I saw for 6.12(d), but I'll try figuring out (b). $\endgroup$ – Nikitau Dec 8 '16 at 18:59
1
$\begingroup$

Use Lebesgue criterion: $\sqrt{f^2+g^2}\le\lvert f\rvert+\lvert g\rvert$, which is bounded, and the discontinuities of $\sqrt{f^2+g^2}$ are a subset of the union of discontinuities of $f$ and $g$, which is a set of zero-measure.

BTW, it is superfluous to say that $f$ and $g$ are bounded. If they are Riemann integrable on a closed interval, they are automatically bounded.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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