5 ms·

It's not true if we pick a=0 and b=0?

• Smaug123 4y ago
  What's not true? The longest inequality on that page is true for all a, b; the rest of the article is about finding specific a, b to extremise the RHS. (Of course, since a=0 b=0 is not a critical point of the RHS, the subsequent lines are not true when a=0 b=0, since they're only true of a critical point.)
  • yobbo 4y ago
    > The longest inequality on that page is true for all a, b If you agree with this, then you why do also believe some particular a,b proves that ∫f²dx ≥ 4?
    • mbustamanter 4y ago
      The premise is that any real number is non-negative when squared, so integration of a squared real function must yield a non-negative number. What is done next is that we now f(x)^2 is greater than a two-variable expression in a and b, so we try to find a maximum. It does not have to do with any particular values because of convexity or concavity of the two variable expression (e.g. if the minimum of a convex function is postivie, then the whole function is positive).
      • yobbo 4y ago
        Yes. The maximum of the RHS is not the minimum of the LHS.
        • Smaug123 4y ago
          Correct. The minimum of the LHS is at least the maximum of the RHS. This is more than you need.
          • yobbo 4y ago
            What is the proven in the article is that the minimum of the LHS is at most 4 and at least 0.
• tgv 4y ago
  To expand on the other answer: by choosing a=0 and b=0, you show that the integral (the LHS) is greater than our equal to zero. The article then shows there are values that prove the lower limit is 4.
  • 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 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 ago
          There 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 ago
            f=6x-2 is just one example. You would need to show that no other f exists.