4 ms·
It seems we have different interpretations of what ∫f²dx ≥ 4 means. I read it as: no f exists for which ∫f²dx is less than 4.
by yobbo 4y ago
It 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.