Hilbert (1) — Soient ${\displaystyle a_{0},\dots ,a_{n))$ des nombres réels positifs ; alors

${\displaystyle \sum _{i=0}^{n}{\sum _{j=0}^{n}{\frac {a_{i}a_{j)){i+j+1))}\leq \pi \cdot \sum _{i=0}^{n}((a_{i))^{2)).}$

Hilbert (1') — 

${\displaystyle \sum _{i=0}^{n}{\sum _{j=0}^{n}{\frac {a_{i}a_{j)){i+j+1))}\leq \pi \cdot \sum _{i=0}^{n}{\sum _{j=0}^{n}((\frac {(i+j)!}{i!j!)){\frac {a_{i}a_{j)){2^{i+j+1))))))$

Hilbert (2) — On a :

${\displaystyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{\frac {a_{n}b_{m)){n+m))<\ {\frac {\pi }{\sin(\pi /p)))\left(\sum _{n=1}^{\infty }a_{n}^{p}\right)^{1/p}\left(\sum _{m=1}^{\infty }b_{m}^{q}\right)^{1/q},}$

avec ${\displaystyle p>1,\ \ q={\frac {p}{p-1)),\ \ {\frac {1}{p))+{\frac {1}{q))=1,\ \ a_{n},b_{m}\geq 0,}$.

Hilbert (2') — 

${\displaystyle \sum _{i=0}^{n}{\sum _{j=0}^{n}{\frac {a_{i}b_{j)){2i+2j+1))}\leq (n+1)\cdot \sin {\frac {\pi }{2(n+1)))\cdot \left(\sum _{i=0}^{n}((a_{i))^{2))\right)^{\frac {1}{2))\cdot \left(\sum _{j=0}^{n}((b_{j))^{2))\right)^{\frac {1}{2))}$

Hilbert (3) — Soit ${\displaystyle (u_{m})}$ un suite de nombres complexes : si la suite est infinie, on la supose de carré sommable (${\displaystyle \sum _{m}|u_{m}|^{2}<\infty }$). Alors

${\displaystyle \left|\sum _{r\neq s}{\dfrac {u_{r}{\overline {u_{s)))){r-s))\right|\leq \pi \displaystyle \sum _{r}|u_{r}|^{2}.}$

Hilbert (3') — Soient ${\displaystyle a_{0},\dots ,a_{n},b_{0},\dots b_{n))$ des nombres complexes. Alors

${\displaystyle {\Bigg |}\sum _{k,j=0}^{n}{\frac {a_{k}{\overline {b_{j)))){1+j+k)){\Bigg |}\leq \pi {\sqrt {\sum _{p=0}^{n}|a_{p}|^{2))}{\sqrt {\sum _{p=0}^{n}|b_{p}|^{2))}.}$

Hilbert (4) — Soient ${\displaystyle f,g\colon [0,\infty )\to [0,\infty )}$ deux fonctions non identiquement nulles, et soient ${\displaystyle p,q}$ des nombres réels avec ${\displaystyle {\tfrac {1}{p))+{\tfrac {1}{q))=1}$. Alors

${\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }{\frac {f(x)g(y)}{x+y))\,\mathrm {d} x\,\mathrm {d} y<{\frac {\pi }{\sin \left({\frac {\pi }{p))\right)))\cdot \left(\int _{0}^{\infty }{f^{p}(x)}\,\mathrm {d} x\right)^{\frac {1}{p))\cdot \left(\int _{0}^{\infty }{g^{q}(x)}\,\mathrm {d} x\right)^{\frac {1}{q))}$