How to define an adequate inner product to show this inequality?

Question

I'm given a continuous function $f\colon[0,1]\to\mathbb{R}^+$ and a sequence $(x_n)_{n\in\mathbb{N}}$ defined by $x_n=\int_0^1t^nf(t)dt$. I must show that for any two $m,n\in\mathbb{N}$, $x_{n+m}\leq\sqrt{x_{2n}}\sqrt{x_{2m}}$. Now, this is in a class of linear algebra, and the inequality looks really similar to the Cauchy-Schwarz inequality for inner products, and so my first idea was to consider $\mathbb{R}$ as an $\mathbb{R}$-vector space and define $g\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ given by $g(x,y)=\int_0^1t^{x+y}f(t)dt$, and so if this were an inner product, the inequality would then follow from the Cauchy-Schwarz one. However, I think from what I've seen, that this is far from being an inner product, and so I have no idea how to proceed, as this is what seemed the more "natural" way of defining one.

Any kind of support is really appreciated. Thanks.

Answer 1

You were definitely on the right path. All you needed is a vector space.

• If $f\equiv 0$, then there's nothing to show.
• Otherwise, let ${\cal P}$ be the vector space of polynomials ${\cal P}$ (with respect to pointwise addition and scalar multiplication), and define

$$(p,q) = \int_0^1 p (t) q(t) f (t) dt, \quad p,q \in{\cal P}.$$

This is an inner product over ${\mathbb R}$, and the result then follows from C-S for $p(t)=t^m$ and $q(t)=t^n$.