Show that: sum from p,q=1 to infinity (x_px_q)/(p+q+1) >
Show that: $$ \sum_{p,q=1}^{\infty} \frac{x_px_q}{p+q+1} > 0 $$ if $0<\sum_{p=1}^{\infty}x_p^2<\infty$.
Explanation of Method
The integral identity $\frac1{p+q+1}=\int_0^1t^{p+q}dt$ converts the double sum into the integral of a perfect square, $\int_0^1\left(\sum_px_pt^p\right)^2dt$, which is manifestly non-negative. Strict positivity then follows because the power series $f(t)=\sum x_pt^p$ cannot vanish identically on $[0,1)$ unless every coefficient $x_p$ is zero, contradicting the hypothesis.
- Represent $1/(p+q+1)$ as an integral and interchange sum and integral.
- Recognize the double sum as $\int_0^1f(t)^2dt$ for $f(t)=\sum_px_pt^p$.
- Show strict positivity via the uniqueness of power series coefficients.
Step 1 · Integral representation
$\dfrac1{p+q+1}=\displaystyle\int_0^1t^{p+q}dt$, so $$ \sum_{p,q=1}^{\infty}\frac{x_px_q}{p+q+1} = \sum_{p,q=1}^{\infty}x_px_q\int_0^1t^{p+q}dt. $$ Since $\sum_p|x_p|t^p\le\left(\sum_px_p^2\right)^{1/2}\left(\sum_pt^{2p}\right)^{1/2}<\infty$ for each fixed $t\in(0,1)$ (Cauchy–Schwarz), the double series converges absolutely for each $t$, justifying the interchange of sum and integral (Fubini's theorem).
Step 2 · Recognize a perfect square
$$ \sum_{p,q=1}^{\infty}\frac{x_px_q}{p+q+1} = \int_0^1\sum_{p,q=1}^{\infty}x_px_qt^{p+q}dt = \int_0^1\left(\sum_{p=1}^{\infty}x_pt^p\right)^2dt = \int_0^1f(t)^2dt \ge 0. $$
Step 3 · Strict positivity
Suppose the integral were $0$. Since $f(t)^2\ge0$ is continuous on $[0,1)$, this forces $f(t)=0$ for all $t\in[0,1)$. But $f(t)=\sum_px_pt^p$ is a power series, so by uniqueness of power-series coefficients, $f\equiv0$ on $[0,1)$ implies every $x_p=0$ — contradicting $\sum_px_p^2>0$. Hence the integral, and the original double sum, is strictly positive.
Summary Table
| Step | Result |
|---|---|
| Double sum | $=\int_0^1f(t)^2dt$, $f(t)=\sum x_pt^p$ |
| Non-negativity | Immediate (perfect square) |
| Strict positivity | Power-series uniqueness rules out $f\equiv0$ |
Additional Notes
- The matrix $H_{pq}=1/(p+q+1)$ is the (shifted) Hilbert matrix, and this problem shows it is positive definite on $\ell^2$ — the quadratic-form analogue of Hilbert's classical inequality bounded in Problem 160.
- The integral-representation trick $1/(p+q+1)=\int_0^1t^{p+q}dt$ is the standard device for turning any Hilbert-type double sum into an integral of a squared (or bilinear) power series, and is reused directly in Problems 159 and 160.
- This is a genuinely infinite-dimensional analogue of the elementary fact that a real symmetric matrix with entries $1/(i+j)$ (finite Hilbert matrices) is positive definite — a classical result in numerical linear algebra, here extended to the full $\ell^2$ setting.
Result
The double sum is strictly positive: