Minimum of integral

Question

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 $$\min\limits_{f\in S}\int\limits_0^1f^2(x)dx.$$

Answer 1

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 $f(x) = 6x$

Answer 2

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).

Answer 3

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.