= Solution
Use the standard $L_\infty$-oriented partition normalization of <B-splines>, in contrast to the unit-integral normalization. Thus $N_i\ge0$ and $\sum_iN_i(t)\le1$ on $[0,1]$, with equality for a complete <basis> on its basic knot interval. This terminology does not mean that each <basis> function's maximum is exactly one; for example an order-three cardinal spline has maximum $3/4$. Assume the <basis> has positive support on the integration interval and is continuous, as is needed for a projection taking values in $C[0,1]$.
Let
$$
H_{ij}=\int_0^1N_i(t)N_j(t)\,dt,\qquad d_i=\int_0^1N_i(t)\,dt>0,\qquad D=\operatorname{diag}(d_i).
$$
The ordinary <Gram matrix> $H$ is positive definite by <linear independence>. The appropriate row-normalized, or <mixed-normalization spline Gram matrix>, is
$$
\boxed{G=D^{-1}H,\qquad G_{ij}=\int_0^1M_i(t)N_j(t)\,dt,\qquad M_i=N_i/d_i.}
$$
Here $M_i$ is a unit-integral nonnegative spline. Unlike $H$, $G$ need not be symmetric. If the full support lies inside the interval, $d_i=(t_{i+k}-t_i)/k$; otherwise use the actual interval integral rather than that full-support formula.
The <normal equations> for the $L_2$ <orthogonal projection> $P_{\mathcal S}f=\sum_i a_iN_i$ say $Ha=c$, with $c_i=\int fN_i$. Divide each row by $d_i$ to obtain
$$
Ga=b,\qquad b_i=\int_0^1f(t)M_i(t)\,dt.
$$
Nonnegativity and unit integral give $|b_i|\le\|f\|_\infty$. The <subpartition of unity for B-splines> then gives
$$
|P_{\mathcal S}f(t)|\le\sum_i|a_i|N_i(t)\le\|a\|_{\ell^\infty}\le\|G^{-1}\|_{\ell^\infty}\|f\|_\infty.
$$
Taking the supremum over $t$ and then over nonzero $f$ proves the <uniform norm bound for B-spline orthogonal projection>:
$$
\boxed{\|P_{\mathcal S}\|_{C[0,1]\to C[0,1]}\le\|G^{-1}\|_{\ell^\infty}.}
$$
The row normalization is essential; replacing $G$ with the ordinary <Gram matrix> would not be the same asserted estimate. The operator is defined on $C[0,1]$ and maps into its finite-dimensional spline subspace, \b[onto $\mathcal S$, not onto all of $C[0,1]$] as one phrase in the PDF says. Its range follows both from the formula and from reproducing every $s\in\mathcal S$.
If “$L_\infty$-normalized” were instead read literally as individually peak-one splines, their sum can exceed one. The basis-dependent factor $K=\|\sum_iN_i\|_\infty$ must then be retained: $\|P_{\mathcal S}\|_\infty\le K\|(D^{-1}H)^{-1}\|_{\ell^\infty}$. Equivalently the displayed bound without an outside factor uses $G=K^{-1}D^{-1}H$. The standard partition-normalized convention has $K\le1$ and yields the sharper bound proved above.
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