Show that the formula \[ \left(\sum_j a_j x^j \,\middle|\, \sum_k b_k x^k\right) = \sum_{j,k} \frac{a_j b_k}{j + k + 1} \] defines an inner product on the space \(R[x]\) of polynomials over the field \(\RR\). Let \(W\) be the subspace of polynomials of degree less than or equal to \(n\). Restrict the above inner product to \(W\), and find the matrix of this inner product on \(W\), relative to the ordered basis \(\{1, x, x^2, \ldots, x^n\}\). (Hint: To show that the formula defines an inner product, observe that \[ (f \mid g) = \int_0^1 f(t)g(t)\,dt \] and work with the integral.)
Solution
For real polynomials \[ f(x)=\sum_j a_jx^j, \qquad g(x)=\sum_k b_kx^k, \] one has \[ \boxed{(f\mid g)=\int_0^1 f(t)g(t)\,dt.} \]
1. The displayed formula defines an inner product on \(\mathbb R[x]\).
Since the sums are finite, \[ \begin{aligned} \int_0^1f(t)g(t)\,dt &=\int_0^1\sum_{j,k}a_jb_k t^{j+k}\,dt\\ &=\sum_{j,k}a_jb_k\int_0^1t^{j+k}\,dt\\ &=\sum_{j,k}\frac{a_jb_k}{j+k+1}. \end{aligned} \] Thus the proposed formula is the ordinary integral pairing. Symmetry and bilinearity follow immediately from the integral.
Also \[ (f\mid f)=\int_0^1f(t)^2\,dt\ge0. \] If this integral is zero, continuity and nonnegativity of \(f^2\) imply \(f(t)=0\) for every \(t\in[0,1]\). A polynomial that vanishes on an interval is the zero polynomial. Hence \((f\mid f)>0\) for every nonzero \(f\), proving positive definiteness. Therefore this is an inner product [@HK71].
2. Its matrix on polynomials of degree at most \(n\) is the Hilbert matrix.
Let \[ \mathcal B=(1,x,x^2,\ldots,x^n). \] For \(0\le j,k\le n\), \[ (x^j\mid x^k) =\int_0^1t^{j+k}\,dt =\frac1{j+k+1}. \] Hence the Gram matrix is \[ \boxed{ H_{n+1} =\left(\frac1{j+k+1}\right)_{0\le j,k\le n}.} \] Equivalently, \[ H_{n+1}= \begin{pmatrix} 1&\frac12&\frac13&\cdots&\frac1{n+1}\\ \frac12&\frac13&\frac14&\cdots&\frac1{n+2}\\ \vdots&\vdots&\vdots&&\vdots\\ \frac1{n+1}&\frac1{n+2}&\frac1{n+3}&\cdots&\frac1{2n+1} \end{pmatrix}. \]