5 ms·
> The article claims the RHS maximum is the minimum for the LHS. No, it shows that the LHS is at least as large as the RHS for any reals a and b (1), then show
by Someone 4y ago
> The article claims the RHS maximum is the minimum for the LHS.
No, it shows that the LHS is at least as large as the RHS for any reals a and b (1), then shows reals a and b exist that make the RHS equal to 4 (2), and from (1) and (2) concludes the LHS is at least as large as 4.
I don’t see it claim that that bound is tight, and the LHS can be made equal to 4.
- yobbo 4y agoIt seems we have different interpretations of what ∫f²dx ≥ 4 means. I read it as: no f exists for which ∫f²dx is less than 4.
- tgv 4y agoThere might be a higher lower bound. So for all f that meet the requirements, ∫f(x)²dx ≥ d, the article shows that d is at least 4. I can show that d is precisely 4, because that's the value for f(x) = 6x - 2.
- yobbo 4y agof=6x-2 is just one example. You would need to show that no other f exists.