I saw this problem and it immediatly took me back to my undergrad days in which I had an optimal control class, where one of the topics studied is minimizing expressions of this kind by abstracting the problem to a tool called functional ([0], [1]) and then applying different methods of resolution, a combination of Lagrange multipliers and the Euler-Lagrange equation [2].
In this particular example, we take the expression to be minimized and then sum the restricions scaled by parameters λ1 and λ2. It should look like this (y=f(x)): ∫ y^2+λ1y+λ2xy dx integrated from 0 to 1. Then, we grab what is inside and apply the Euler-Lagrange equation: L(x,y)=y^2+λ1y+λ2xy ⇒ 2y+λ1y+λ2x=0. So now we now f(x) is linear (which is why the method in the post works, if f(x) was a polynomial of higher order, then it should have been substracted a poylynomial in x other than ax+b). Solving for y and substituting in the restrictions, we integrate linear and quadratic functions (easy) to arrive at a system of equations for λ1 and λ2. Solving we get (λ1,λ2)=(4,-12). Then f(x)=6x-2 is the minimizer for the functional. Integrating f(x)^2 we get that its integral is 4, so for any f(x) satisfying those two conditions, ∫ f(x)^2 dx is greater than or equal to 4, and that lower bound is achieved for f(x)=6x-2.
I didn't mention convexity, which is what allows us to conclude that the inequality is in that direction, but I leave it to the reader.
We can also find the f minimizing the integral of f^2 by considering the Lagrangian: integral f(x)^2 - lambda f(x) + lambda - mu x f(x) + mu dx, then solving pointwise for f(x) to get f(x) = (lambda + mu x) / 2, then using the constraints to get lambda and mu.
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.)
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).
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.
> 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.
I saw this problem and it immediatly took me back to my undergrad days in which I had an optimal control class, where one of the topics studied is minimizing expressions of this kind by abstracting the problem to a tool called functional ([0], [1]) and then applying different methods of resolution, a combination of Lagrange multipliers and the Euler-Lagrange equation [2].
In this particular example, we take the expression to be minimized and then sum the restricions scaled by parameters λ1 and λ2. It should look like this (y=f(x)): ∫ y^2+λ1y+λ2xy dx integrated from 0 to 1. Then, we grab what is inside and apply the Euler-Lagrange equation: L(x,y)=y^2+λ1y+λ2xy ⇒ 2y+λ1y+λ2x=0. So now we now f(x) is linear (which is why the method in the post works, if f(x) was a polynomial of higher order, then it should have been substracted a poylynomial in x other than ax+b). Solving for y and substituting in the restrictions, we integrate linear and quadratic functions (easy) to arrive at a system of equations for λ1 and λ2. Solving we get (λ1,λ2)=(4,-12). Then f(x)=6x-2 is the minimizer for the functional. Integrating f(x)^2 we get that its integral is 4, so for any f(x) satisfying those two conditions, ∫ f(x)^2 dx is greater than or equal to 4, and that lower bound is achieved for f(x)=6x-2.
I didn't mention convexity, which is what allows us to conclude that the inequality is in that direction, but I leave it to the reader.
[0] https://en.wikipedia.org/wiki/Functional_(mathematics)
[1] https://en.wikipedia.org/wiki/Functional_derivative
[2] https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equatio...
We can also find the f minimizing the integral of f^2 by considering the Lagrangian: integral f(x)^2 - lambda f(x) + lambda - mu x f(x) + mu dx, then solving pointwise for f(x) to get f(x) = (lambda + mu x) / 2, then using the constraints to get lambda and mu.
I liked this because it uses a method to prove something which seems very disconnected from the initial problem.
It's not true if we pick a=0 and b=0?
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.)
> 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?
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).
Yes. The maximum of the RHS is not the minimum of the LHS.
Correct. The minimum of the LHS is at least the maximum of the RHS. This is more than you need.
What is the proven in the article is that the minimum of the LHS is at most 4 and at least 0.
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.
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.
> 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.
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.
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.
f=6x-2 is just one example. You would need to show that no other f exists.