4 ms·
The article claims the RHS maximum is the minimum for the LHS. > ... you show that the integral (the LHS) is greater than our equal to zero. The article adds
by yobbo 4y ago
The article claims the RHS maximum is the minimum for the LHS.
> ... you show that the integral (the LHS) is greater than our equal to zero.
The article adds a function g, and uses (f+g)²≥0. It then picks a random g (g=ax+b), such that it can be simplified after expanding.
- 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.