Problem 1452

To prove that
$$\int_0^1 f(x)^2 \, dx \ge n^2,$$
we will use the properties of orthonormal polynomials and the given conditions.

1. Legendre Polynomials and Orthonormal Basis:
Let $\{P_k\}_{k=0}^{\infty}$ denote the sequence of Legendre polynomials. These polynomials form an orthonormal basis for $L^2([0,1])$ when appropriately scaled. Specifically, let $Q_k = \sqrt{2k+1} P_k$, then $\{Q_k\}_{k=0}^{\infty}$ form an orthonormal basis for $L^2([0,1])$.

2. Inner Product and Orthonormality:
For any function $f \in L^2([0,1])$, we can express $f$ in terms of the orthonormal basis $\{Q_k\}$:
$$f(x) = \sum_{j=0}^{\infty} \langle f, Q_j \rangle Q_j(x),$$
where $\langle f, Q_j \rangle$ denotes the inner product:
$$\langle f, Q_j \rangle = \int_0^1 f(x) Q_j(x) \, dx.$$

3. Given Conditions:
The problem states that for every $k \in \{0, 1, \ldots, n-1\}$,
$$\int_0^1 x^k f(x) \, dx = 1.$$
Since $P_k(1) = 1$ for Legendre polynomials, we have:
$$\int_0^1 P_k(x) f(x) \, dx = 1.$$

4. Norm Calculation:
Using the orthonormality of $\{Q_k\}$, we can write:
$$\int_0^1 f(x)^2 \, dx = \sum_{j=0}^{\infty} |\langle f, Q_j \rangle|^2.$$
Since $\langle f, Q_j \rangle = \sqrt{2j+1} \int_0^1 P_j(x) f(x) \, dx$, for $j \le n-1$, we have:
$$\langle f, Q_j \rangle = \sqrt{2j+1}.$$
Therefore,
$$\int_0^1 f(x)^2 \, dx \ge \sum_{j=0}^{n-1} |\langle f, Q_j \rangle|^2 = \sum_{j=0}^{n-1} (2j+1).$$

5. Summation:
The sum of the first $n$ odd numbers is given by:
$$\sum_{j=0}^{n-1} (2j+1) = n^2.$$

Thus, we have:
$$\int_0^1 f(x)^2 \, dx \ge n^2.$$

$\blacksquare$