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Which is the smallest constant $B_d$ such that the following inequality $$\left|\int_{0}^{1}t^d(1-t)\psi(t) dt\right|^2\le B_d\int_{0}^{1}t^d|\psi(t)|^2dt$$ holds, provided that $\psi(t)$ is a polynomial?

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up vote 0 down vote accepted

A hint: Consider the scalar product $$\langle u,v\rangle:=\int_0^1 u(t) v(t)\ t^d\ dt\ .$$

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Presumably $d > -1$ so the integrals converge.

Consider the Hilbert space $L^2(\mu)$ where $\mu$ is the measure $\mu(dt) = t^d \ dt$ on $[0,1]$. Your inequality says $|\langle 1-t, \psi\rangle|^2 \le B_d \|\psi\|^2$. The Cauchy-Schwarz inequality says the inequality is true for all $\psi \in L^2(\mu)$, in particular for polynomials, with $$B_d = \|1-t\|^2 = \int_0^1 t^d (1-t)^2\ dt = \dfrac{2}{(d+1)(d+2)(d+3)}$$ and this is best possible (since you could take $\psi(t) = 1-t$).

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