Integrated Legendre projection gives an explicit p-rate in H1

By [F2], finite Legendre sums span all polynomials. The density facts [F8] and [F9] imply that polynomials are dense in $L^2(-1,1)$: first approximate an $L^2$ class by a compactly supported smooth function, then approximate that function uniformly on $[-1,1]$ by a polynomial. The coefficient norm at $s=0$ is the ordinary $L^2$ norm on finite Legendre sums by [F2]–[F3], and [F10] makes $L^2(-1,1)$ complete. Hence the identity on finite sums extends to an isometry from $H_L^0$ onto $L^2(-1,1)$. Since each spectral weight for $s\ge 0$ is at least its $s=0$ weight, [F3] gives a continuous embedding $H_L^s\to H_L^0$. Thus $g$ has the canonical $L^2$ class in the Statement, and Cauchy–Schwarz [F7] on the interval of length two gives $g\in L^1$. Then [F5] makes $u$ absolutely continuous. The definition [F1] makes $I_{p}u$ the integral of a polynomial of degree at most $p-1$, hence a degree-at-most-$p$ polynomial. By the finite projection formula in [F1] and Legendre orthogonality [F2], $g-{\mathrm{\Pi }}_{p-1}g$ is orthogonal to constants, so $\int {\mathrm{\Pi }}_{p-1}g=\int g$ and the endpoint values agree. The primitive defining $I_{p}u$ is absolutely continuous by [F5].

Apply [F6] to the two primitives in step 1.1; their a.e. derivatives are $g$ and ${\mathrm{\Pi }}_{p-1}g$, so $e=u-I_{p}u$ has derivative $e^{\mathrm{\prime }}=g-{\mathrm{\Pi }}_{p-1}g$. By [F3] the finite projection is the truncation in [F1]; the tail estimate [F4] with $N=p-1$ gives $\mathrm{\parallel }{e}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}\le (1+p(p+1){)}^{-s\mathrm{/}2}\mathrm{\parallel }g{\mathrm{\parallel }}_{H_L^s}$. Since $e(-1)=0$, $e(x)={\int }_{-1}^x{e}^{\mathrm{\prime }}(t)dt$; Cauchy–Schwarz [F7] and integration over an interval of length two give $\mathrm{\parallel }e{\mathrm{\parallel }}_{L^2}\le 2\mathrm{\parallel }{e}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}$. Therefore $\mathrm{\parallel }e{\mathrm{\parallel }}_{H^1}^2=\mathrm{\parallel }e{\mathrm{\parallel }}_{L^2}^2+\mathrm{\parallel }{e}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2\le 5\mathrm{\parallel }{e}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2$, proving the estimate.