= Solution
There are $n$ basis <splines> for the stated $n+k$ knots; the upper limit $k$ in the printed basis list is an indexing slip. Let
$$
A_{ij}=N_j(x_i^*),\qquad i,j=1,\ldots,n.
$$
The strictly increasing sites and $t_i<x_i^*<t_{i+k}$ give invertibility by the <Schoenberg–Whitney theorem>. The needed additional <matrix> property is <total nonnegativity of B-spline collocation matrices>: every ordered minor of $A$ is nonnegative.
Here is a structural justification of that sign property. <Knot insertion> refines <spline> coefficients by rules of the form $b_j=\alpha_j a_j+(1-\alpha_j)a_{j-1}$, $0\leq\alpha_j\leq1$, with the usual unchanged coefficients before the insertion and shifted coefficients after it. Each coefficient map is a nonnegative rectangular bidiagonal <matrix>; all its ordered minors are nonnegative, as its nonzero terms preserve row/column order. Products preserve this property by the <Cauchy–Binet formula>. Insert each ordered sampling site until it has full knot multiplicity. Evaluation at such a break is one of the refined coefficients, with a consistent one-sided convention. Consequently the <B-spline collocation matrix> is an ordered row submatrix of the product of refinement <matrices>, proving its nonnegative minors. This also applies to repeated admissible original knots.
Since $A$ is invertible and totally nonnegative, $\det A>0$. The <adjugate identity> gives the <checkerboard inverse of a totally nonnegative matrix>:
$$
(A^{-1})_{ij}=(-1)^{i+j}\frac{\det A_{\widehat j,\widehat i}}{\det A},
\qquad (-1)^{i+j}(A^{-1})_{ij}\geq0.
$$
Now for any <spline> $s=\sum_j a_jN_j$, its sampled-value vector is $y=Aa$. Hence the coefficient <linear functional> in the <dual basis> is exactly
$$
\boxed{\mu_i(s)=a_i=\sum_j(A^{-1})_{ij}s(x_j^*).}
$$
In particular $\mu_i(N_\ell)=\delta_{i\ell}$, verifying the hinted formula rather than assuming it. The <operator norm> of this <linear functional> is taken for the <uniform norm> on the <spline>'s interval, containing all sampling sites. Evaluation at a sampling site has norm at most one, so
$$
|\mu_i(s)|\leq\left(\sum_j|(A^{-1})_{ij}|\right)\|s\|_\infty,
\qquad
\|\mu_i\|\leq\sum_j|(A^{-1})_{ij}|.
$$
For the given unit-norm alternating <spline>, $s_*(x_j^*)=(-1)^j$, and the checkerboard signs make all terms in a fixed inverse row align:
$$
a_i^*=\mu_i(s_*)=\sum_j(A^{-1})_{ij}(-1)^j
=(-1)^i\sum_j|(A^{-1})_{ij}|.
$$
Since $\|s_*\|_\infty=1$, this supplies the reverse <operator norm> inequality and proves the <Chebyshev spline coefficient and dual norm equality>
$$
\boxed{|a_i^*|=\|\mu_i\|=\sum_j|(A^{-1})_{ij}|,\qquad i=1,\ldots,n.}
$$
This <spline> attains the <operator norm> of each coefficient <linear functional>.
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