Let $k \in \{0,1,...,n-1 \}$ and $f:[0,1] \to \mathbb{R}$ be a continous function. If $\int_{[0,1]} x^k f(x) dx =1$ for all such $k$ then show that $\int_{[0,1]} (f(x))^2 dx \ge n^2$.


Let $\{A_n(x)\}_{n\in\mathbb{N}}$ be the sequence of the shifted Legendre polynomials, $A_n(x)=P_n(2x-1)$.

This sequence gives an orthogonal base of $L^2([0,1])$ with respect to the dot product $\langle f,g\rangle = \int_{0}^{1} f(x)g(x)\,dx $:

$$\int_{0}^{1} A_n(x)\,A_m(x)\,dx = \frac{\delta_{n,m}}{2n+1}\tag{1}$$ and for every $n\in\mathbb{N}$ we have that $A_n(x)$ is a degree-$n$ polynomial such that $A_n(1)=1$.

By orthogonality, for every $k\leq n-1$: $$ \int_{0}^{1} f(x)\, A_k(x)\,dx = A_k(1) = 1 \tag{2}$$ hence if we decompose $f(x)$ as: $$ f(x) = \sum_{h\geq 0} a_h A_h(x) \tag{3}$$ we have $a_h=2h+1$ for every $h\leq n-1$ and: $$ \int_{0}^{1}f(x)^2\,dx \geq \sum_{h=0}^{n-1}\frac{a_h^2}{2h+1} = \sum_{h=0}^{n-1}(2h+1)=\color{red}{n^2}\tag{4}$$ as wanted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.