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Let $S$ be the set of all integrable on $[0,1]$ such that $$\int\limits_0^1f(x)dx=\int\limits_0^1xf(x)dx+1=3.$$ Prove that $S$ is infinite and evaluate $$\min\limits_{f\in S}\int\limits_0^1f^2(x)dx.$$

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To show that $S$ is infinite, just find two elements of $S$; then any linear combination is also a solution. For the minimum, the Euler-Lagrange equations with Lagrange multipliers say that $f$ should be linear, and you can determine the coefficients from the conditions; then you need to show that this is minimal in the absence of the differentiability assumptions of the Euler-Lagrange equations. – joriki Mar 16 '12 at 7:39
To show that the solution $f_0$ of the Euler-Lagrange equation is minimal even in absence of the differentiability assumption, write $f = f_0 + g$ and minimize $\int f(x)^2\,dx$ with respect to $g$ under the condition $\int g(x)\, dx= \int x g(x)\, dx =0$. – Fabian Mar 16 '12 at 9:10

We have $3 = \int_0^1 f(x) \cdot \left(x +\frac{1}{3} \right) \, dx$. Let's use Cauchy–Schwarz inequality:

$$3 = \int_0^1 f(x) \cdot \left(x +\frac{1}{3} \right) \, dx \le \sqrt{ \int_0^1 f^2(x) \, dx} \sqrt{\int_0^1 \left(x +\frac{1}{3} \right)^2 dx} = \sqrt{\frac{7}{9}} \sqrt{ \int_0^1 f^2(x) \, dx}$$ Hence: $$\int_0^1 f^2(x) \, dx \ge \left( \frac{9 \cdot 3}{\sqrt{7}} \right)^2 = \frac{81}{7} \approx 11.57$$


$$2 = \int_0^1 x f(x) \, dx \le \sqrt{ \int_0^1 f^2(x) \, dx} \sqrt{\int_0^1 x^2 dx} = \frac{1}{\sqrt{3}} \sqrt{ \int_0^1 f^2(x) \, dx}$$ So: $$\int_0^1 f^2(x) \, dx \ge 12 $$ And equality holds if $f(x) = 6x$

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Can you give us a function for which this minimum is achieved? I think your bound is wrong. I am pretty sure that the disired minimum equals 12. – passenger Jun 18 '12 at 18:02
@passenger sorry, I edited my post and now bound is a bit less than 12. Equality in C-S holds if $f(x) = A(x+\frac{1}{3})$ but unfortunately there is no $A$ such that $f$ satisfies given conditions. – qoqosz Jun 18 '12 at 18:17
I found the same! – passenger Jun 18 '12 at 19:28

Assuming $f\in L^1([0,1])\cap L^2([0,1])$, we can take an expansion of $f$ in terms of the shifted Legendre polynomials: $$ f(x) = a_0+\sum_{n\geq 1} c_n P_n(2x-1).\tag{1} $$ Since this set of polynomials is an orthogonal base of $L^2([0,1])$: $$ \int_{0}^{1} P_n(2x-1) P_m(2x-1)\,dx = \frac{\delta_{m,n}}{2n+1}\tag{2} $$ and $P_0(2x-1)=1,P_1(2x-1)=2x-1$, the given constraints translate into: $$ a_0 = \int_{0}^{1}f(x)\,dx = 3, $$ $$ a_1 = 3\int_{0}^{1}(2x-1)f(x)\,dx = -9+6\int_{0}^{1}x\,f(x)\,dx =3.\tag{3}$$ So we have that any function of the form $$ f(x) = 6x + \sum_{n\geq 2}c_n P_n(2x-1) $$ satisfies the given constraints, and: $$ \int_{0}^{1}f(x)^2\,dx = a_0^2+\sum_{n\geq 1}\frac{c_n^2}{2n+1} \geq a_0^2+\frac{a_1^2}{3},\tag{4}$$ so the minimum for the LHS, $12$, is attained by $f(x)=6x$.

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Looking at the polynomial $ax^{n+1}+bx^n$, from the equations we can derive that the coefficients satisfy two linear equations, yielding the following system:

$$\begin{pmatrix} \frac{1}{n+2} & \frac{1}{n+1} \\\frac{1}{n+3} & \frac{1}{n+2}\end{pmatrix} \cdot \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$$

The determinant, $\frac{-1}{(n+1)(n+2)^2(n+3)}$, is non-zero for all $n \in \mathbb{N}$, so we have directly found infinite polynomial solutions. This of course, is not to say that these are all the solutions, so it doesn't necessarily help in evaluating the minimum (or even proving that one exists).

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A hint: For any given function $g:\ [0,1]\to{\mathbb R}$ there is an $f$ of the form $$f(x):=a + b x + g(x)$$ that fulfills the given conditions. This already proves the first part. Furthermore, any $f$ fulfilling the given conditions can be written in the above form with certain $a$, $b$ and $$\int g(x)\ dx=0,\qquad \int_0^1 x\,g(x)\ dx=0\ .$$ Use this to solve the minimum problem.

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