Minimum of integral 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 $$\min\limits_{f\in S}\int\limits_0^1f^2(x)dx.$$
 A: 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:
$$\int_0^1 f^2(x) \, dx \ge \left( \frac{9 \cdot 3}{\sqrt{7}} \right)^2 = \frac{81}{7} \approx 11.57$$
Edit
$$2 = \int_0^1 x f(x) \, dx \le \sqrt{ \int_0^1 f^2(x) \, dx} \sqrt{\int_0^1 x^2 dx} = \frac{1}{\sqrt{3}} \sqrt{ \int_0^1 f^2(x) \, dx}$$
So:
$$\int_0^1 f^2(x) \, dx \ge 12 $$
And equality holds if $f(x) = 6x$
A: 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 $f(x)=6x$.
A: 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).
A: 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.
