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For $f$ of degree $d$ and $n\ge \max (1,d)$, compute its degree-$n$ Bernstein basis coefficients and solve $\max \lambda$ subject to $\lambda \le {\beta }_{k,n}$, $0\le k\le n$. This is a linear program with $n+1$ inequalities. The coefficients form a convex-combination enclosure of the polynomial values, so $v_n\le \min f$. Degree elevation of Bernstein coefficients makes the bounds monotone. For every $ϵ>0$, the polynomial $f-(\min f-ϵ)$ is strictly positive, so positive Bernstein coefficients for a strictly positive polynomial gives an eventual feasible bound $\min f-ϵ$. Hence $v_n$ converges to the true minimum. Earlier indices $n<d$ can use a one-inequality program at any common coefficient-based lower bound, preserving the stated constraint count for every positive index.

Use degree elevation of Bernstein coefficients. If $b_j$ are degree-$p$ Bézier curve controls, the identical curve has degree-$(p+1)$ controlsThis follows by multiplying each degree-$p$ Bernstein basis polynomial by $(1-t)+t$ and collecting the degree-$(p+1)$ terms, so the operation changes the representation, not the curve.

For the quadratic controls in part (a), the cubic controls areRaise these cubic Bézier curve controls once more. ThenThus the requested quartic Bézier curve controls, expressed in the original polygon, areEach expression is an affine combination whose coefficients sum to one. Endpoint positions and derivatives are preserved at each degree-raising step, as is every point of the span.

Use the quartic treble-knot representation of a quadratic B-spline. Specify the indexing as well as the controls: for the interior of the long curve, take the quartic spline knot sequenceWith the standard B-spline convention that control $Q_j$ multiplies $N_{j,4}$ supported on $[{\tau }_j,{\tau }_{j+5}]$, the answer isIn other words, keep the three interior quartic Bézier curve controls $E_{1,i},E_{2,i},E_{3,i}$ for each span in that order. The offset $3i-1$ matters for the stated knot indexing; shifting all indices together gives an equivalent convention.

To prove the representation, insert each integer interior knot once, changing multiplicity three to four. A degree-four knot of multiplicity four separates Bézier curve spans while leaving their shared endpoint common. The three interior controls $E_{1,i},E_{2,i},E_{3,i}$ remain, and knot insertion constructs the shared endpoint from its neighboring controls with weights $1/2,1/2$, since adjacent knot intervals have equal length. In fact,The right endpoint is obtained in the same way. Hence every extracted quartic span has exactly the five controls proved in part (b). Knot insertion leaves the curve unchanged, so the assembled quartic B-spline is identical to the original quadratic one on every span.

There is also a derivative check at each join:This is why treble knots are appropriate: degree four with interior knot multiplicity three permits $C^{4-3}=C^1$ continuity, exactly the generic continuity of the uniform quadratic B-spline. For a finite clamped curve, include its endpoint controls and endpoint knot multiplicities using the same degree elevation of Bernstein coefficients and knot insertion; the displayed formulas describe the long interior requested here, without imposing an unstated end condition.

Let $f(x)={\sum }_{j=0}^d{a}_j{x}^j$. For $n\ge \max (1,d)$, express $f$ in the Bernstein basis as $f={\sum }_{k=0}^n{\beta }_{k,n}{b}_{k,n}$, where $b_{k,n}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^k(1-x{)}^{n-k}$. Its precomputed coefficients arewith the ratio for $j=0$ interpreted as one. The identity follows from $x^j={\sum }_k[(k{)}_j/(n{)}_j]b_{k,n}$, obtained by the binomial theorem.

At this level solve the linear programThere are exactly $n+1$ inequalities. Equivalently, $f-\lambda$ must have nonnegative coefficients in the unnormalized basis; its coefficients are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)({\beta }_{k,n}-\lambda )$. Since the basis functions are nonnegative and sum to one, every feasible $\lambda$ satisfies $\lambda \le f(x)$ for all $x\in [0,1]$, so $v_n\le f_{\min }$.

Degree elevation of Bernstein coefficients gives, for $1\le k\le n$,with endpoint coefficients unchanged. These convex combinations imply $v_{n+1}\ge v_n$.

To cover every positive index even when $d>1$, put $\ell =a_0-{\sum }_{j=1}^d|a_j|$. For $1\le n<d$, use the one-inequality LP $\max \{\lambda :\lambda \le \ell \}$, giving $v_n=\ell$. This is a lower bound on $f$; moreover the explicit coefficient formula has all its factorial ratios in $[0,1]$, so ${\beta }_{k,d}\ge \ell$. Thus the transition to level $d$ is also monotone and every level respects the requested bound of at most $n+1$ inequalities. Constants are handled by the main formula at all positive indices.

Finally, for every $ϵ>0$, $f-(f_{\min }-ϵ)$ is strictly positive on the interval. Part (c) supplies a nonnegative-coefficient representation at some degree, and degree elevation preserves it at every higher degree. Hence eventually $v_n\ge f_{\min }-ϵ$. Combining with the upper bound proves the Bernstein linear programming hierarchy for polynomial minimization:

For a uniform quadratic B-spline with interior control points $P_i$, label its $i$th span $[i,i+1]$. A quartic B-spline with interior spline knot sequence ${\tau }_{3i}={\tau }_{3i+1}={\tau }_{3i+2}=i$ represents the identical curve whenTo verify this, raise each quadratic Bézier curve span twice by degree elevation of Bernstein coefficients. The three interior quartic controls are the displayed points. One additional knot insertion at each interior knot supplies the shared endpoint as the mean of its two neighboring controls, recovering all five quartic Bézier curve controls. Thus the knot multiplicity is three, giving $C^1$ joins rather than imposing unwanted extra smoothness.