Minimize $\int\limits _{0}^{1} f^2(x) dx$ if $\int\limits_{0}^{1} f(x) dx = 1$, $\int\limits_{0}^{1} xf(x)dx = 1$

Question

Any hints? Using integration by parts didn't produce any result. Interestingly, I cannot find any function $f(x)$ which such a property.

Answer 1

Thank you user159517 ... it is an Euler-Lagrange problem ... \begin{eqnarray*} L(x,f(x),f^{'}(x))=(f(x))^2+\lambda f(x)+\mu x f(x) \end{eqnarray*} The Euler-Lagrange equation \begin{eqnarray*} \frac{ \partial L}{\partial f}=\frac{d}{dx} \frac{ \partial L}{\partial f^{'}} \end{eqnarray*} & this gives \begin{eqnarray*} 2f(x)+ \lambda x+\mu=0 \end{eqnarray*} Now the constraints give
\begin{eqnarray*} f(x)=-2+6x \end{eqnarray*} With this I make the minimal value $4$.

Answer 2

Using the standard inner product on $L^2$, we have $$ \langle a+bx,f(x)\rangle=\int_0^1(a+bx)f(x)\,dx=a+b $$ Also $$ \|a+bx\|^2=\int_0^1(a+bx)^2\,dx=a^2+ab+b^2/3 $$ and, by Cauchy-Schwarz, $$ \int_0^1f(x)^2\,dx\ge\frac{(a+b)^2}{a^2+ab+b^2/3} $$ Setting $t=a/b$, we want to find the maximum of $$ g(t)=3\frac{(t+1)^2}{3t^2+3t+1} $$ Since $$ g'(t)=-3\frac{(t+1)(3t+1)}{(3t^2+3t+1)^2} $$ Note that $g(t)\ge0$, so the maximum is at $t=-1/3$, with $$ g(-1/3)=4 $$ This value is actually attained: $$ \int_0^1 (a(-1+3x))^2\,dx=a^2\int_0^2(1-6x+9x^2)\,dx=a^2(1-3+3)=a^2 $$ so $a=\pm2$; also $$ \int_0^1a(-1+3x)\,dx=a(-1+3/2)=a/2 $$ so we need $a=2$. Finally $$ \int_0^12x(-1+3x)\,dx=1 $$

Answer 3

Here is an approach using adjoints working in $L^2[0,1]$.

Let $A:L^2[0,1] \to \mathbb{R}^2$ be defined by $A f = \int_0^1 \binom{x}{1} f(x) dx$, then the problem becomes $\min \{ \|f\|^2 | Af = \binom{1}{1} \}$. Since $L^2[0,1] = \ker A \oplus {\cal R} A^*$, we see that the minimum norm solution to $Af = \binom{1}{1}$ must lie in ${\cal R} A^*$, that is the problem becomes $\min \{ \|A^*y\|^2 | A A^* y = \binom{1}{1} \} = \|A^* (A A^*)^{-1} \binom{1}{1}\|^2$.

Grinding through the computations: $\langle y, Af \rangle = y_1 \int_0^1 x f(x) dx + y_2 \int_0^1 f(x) dx = \langle x \mapsto x y_1 + y_2, f \rangle$ and so $A^* y (x) = x y_1 + y_2$, and $A A^* = \begin{bmatrix} {1 \over 3} & {1 \over 2} \\ {1 \over 2} & 1 \end{bmatrix}$.

Then $(A A^*)^{-1} \binom{1}{1} = \binom{6}{-2}$ and the minimising $f$ is $f(x)=(A(A A^*)^{-1} \binom{1}{1})(x) = 6x-2$ and $\|A^* (A A^*)^{-1} \binom{1}{1}\|^2 = 4$.

Approach using Lagrange multipliers:

First, we must establish that a solution exists, this follows since we are finding the nearest point to a closed convex set.

If $c(f) = \|f\|^2$, then $c'(f)h = \langle f, h \rangle$, if we let $g_1(f), g_2(f)$ denote the constraints in the title, we have $g_1'(f)h = \langle x \mapsto 1, h \rangle$, and $g_2'(f)h = \langle x \mapsto x, h \rangle$.

Note that $g_1', g_2'$ are linearly independent, hence Lagrange says that a solution satisfies $c'(f)+ \lambda_1 g_1'(f) + \lambda_2 g_2'(f)= 0$, or equivalently, $f(x) = -{1 \over 2} (\lambda_1 + \lambda_2x)$. Since $g_1(f) = g_2(f) = 1$ we have $\lambda_1 = 4, \lambda_2 = -12$.

Answer 4

A more elementary version of copper.hat's $L^2[0,1]$ solution:

We can expand $f$ in basis polynomials $$f(x) = c_0+c_1x+c_2p_2(x) +c_3p_3(x)\ldots$$

where $p_2,p_3,\ldots$ are chosen to be orthogonal to $1$ and $x,$ i.e. $$ \int_0^1p_i(x)dx=\int_0^1xp_i(x)dx = 0$$

and orthogonal to each other by the Gram Schmidt process. (The not so rigorous part of this is you'd need to prove/know/believe this spans $L^2[0,1]$ and is thus a basis.)

Then by orthogonality we have $$ \int_0^1f(x)^2dx = \int_0^1(c_0+c_1x)^2 +\sum_{i=2}^\infty c_i^2\int_0^1p_i(x)^2dx.$$

The constraints are $$ 1=\int_0^1f(x)dx = c_0+\frac{1}{2}c_1 \\ 1 = \int_0^1xf(x)dx = \frac{1}{2}c_0+\frac{1}{3}c_1$$ which can be solved to get $c_0 =-2,$ $c_1 = 6.$ We can see from the positivity of the terms of expansion that the minimal solution obeying $c_0=-2$ and $c_1=6$ will have $c_i=0$ for $i\ge 2.$ So we have $$ f(x) = 6x-2$$ for a minimum value of $4$.