GLL quadrature changes only the top Legendre modes and is p-uniformly equivalent

Define $g(x)=(x^2-1)P_p^{\mathrm{\prime }}(x)$; it vanishes at all GLL nodes and has degree $p+1$. For $p\ge 2$, divide any polynomial $f$ of degree at most $2p-1$ as $f=gr+s$ with $\mathrm{deg}r\le p-2$ and $\mathrm{deg}s\le p$. Integration by parts gives ${\int }_{-1}^{1}gr=-{\int }_{-1}^1{P}_p(2xr+(x^2-1)r^{\mathrm{\prime }})=0$, since the boundary term vanishes and the bracket has degree at most $p-1$. The quadrature of $gr$ is also zero because $g$ vanishes at every node, while the interpolatory definition integrates $s$ exactly. Thus $Q_p(f)=\int f$. For $p=1$, interpolatory exactness through degree $p=1$ is the required degree $2p-1=1$.

Divide $P_p^2=gr+s$ with $\mathrm{deg}r=p-1$ and $\mathrm{deg}s\le p$. Since the leading coefficient of $g$ is $p{\kappa }_p$, the leading coefficient of $r$ is ${\kappa }_p\mathrm{/}p$. Integration by parts as in step 1.1 gives $\int gr=-\int P_p(2xr+(x^2-1)r^{\mathrm{\prime }})$. The bracket has degree at most $p$ and leading coefficient $(p+1){\kappa }_p\mathrm{/}p$, so it equals $(p+1)P_p\mathrm{/}p$ plus a polynomial of degree below $p$; orthogonality [F2] therefore gives $\int gr=-(p+1)\mathrm{\parallel }{P}_p{\mathrm{\parallel }}_2^2\mathrm{/}p$. Since $Q_p(gr)=0$ and $Q_p(s)=\int s$, it follows that $Q_p(P_p^2)-\int P_p^2=-\int gr=(p+1)\mathrm{\parallel }{P}_p{\mathrm{\parallel }}_2^2\mathrm{/}p=2(p+1)\mathrm{/}(p(2p+1))$.

Write any degree-$p$ polynomial as $f=a_p{P}_p+r$ with $\mathrm{deg}r<p$. Step 1.1 integrates $r^2$ and $P_{p}r$ exactly, so step 2.1 gives $Q_p(f^2)=\int f^2+{\delta }_p{a}_p^2$, where ${\delta }_p=2(p+1)\mathrm{/}(p(2p+1))>0$. For a cardinal polynomial ${\mathrm{\ell }}_j$, $Q_p({\mathrm{\ell }}_j^2)=w_j$ by its nodal values, while the displayed identity gives $w_j=\int {\mathrm{\ell }}_j^2+{\delta }_p{a}_p({\mathrm{\ell }}_j{)}^2>0$. Thus all quadrature weights are positive and the one-dimensional squared norm ratio is between $1$ and $2+1\mathrm{/}p\le 3$.

Expand a tensor polynomial in the orthogonal basis ${\prod }_{r=1}^d{P}_{{\nu }_r}({\xi }_r)$, $0\le {\nu }_r\le p$. The tensor quadrature and exact integral are both diagonal on this basis: every cross term is exact by step 1.1, and each diagonal factor has ratio $1$ unless ${\nu }_r=p$, when step 3.1 bounds it by $3$. Multiplying the $d$ coordinate factors proves the mass norm bounds. Each coordinate derivative of a tensor polynomial is still in ${\mathbb{Q}}_p$, so applying the same tensor norm bounds to each component and summing proves the gradient bounds.

An affine change of variables multiplies both exact integrals and the transported quadrature by the same positive Jacobian. Apply the unweighted bounds of step 4.1 to each physical gradient component pulled back to the reference cube. The measurable bounded coefficient makes $a\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2$ integrable on the bounded box, and its specified values on $\bar{K}$ define every quadrature sample. Positivity of the weights and the pointwise bounds give $\underset{‾}{a}Q(\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2)\le Q(a\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2)\le \bar{a}Q(\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2)$ and the analogous inequalities for the exact integral; combine them with step 4.1 to obtain the stated weighted comparison. The constants are independent of $p$.