Analytic derivative bounds give exponential polynomial approximation on a triangle

For $0<\tau <1$ and integer $n\ge 1$ define $B_{n,\tau }(t)={\sum }_{j=\lceil n\tau \rceil }^n\left(\genfrac{}{}{0px}{}{n}{j}\right)t^j(1-t{)}^{n-j}$. Then $0\le B\le 1$, $\mathrm{deg}B\le n$, and $\mathrm{\mid }{B}^{\mathrm{\prime }}\mathrm{\mid }\le n$. If $t\le \tau -\delta$ for fixed $\delta >0$, Chernoff's bound follows directly from $1_{j\ge n\tau }\le e^{z(j-n\tau )}$ for $z>0$: $B_{n,\tau }(t)\le e^{-zn\tau }(1-t+t{e}^z{)}^n\mathrm{.}$ For sufficiently small fixed $z$, the exponent per $n$ is negative uniformly because its derivative at $z=0$ is $t-\tau \le -\delta$. Applying the same argument to the complementary lower tail gives exponential smallness of $1-B$ on $t\ge \tau +\delta$. Differentiation telescopes to one binomial atom times $n$; away from the threshold its probability is bounded by the same tail estimate with gap $\delta \mathrm{/}2$ for all sufficiently large $n$, and increasing the constant covers the finitely many smaller $n$. Thus $B$ and $1-B$ switch from exponentially small to $1+O(e^{-cn})$ across a strip of fixed width, and their derivatives outside that strip are $O(n{e}^{-cn})$.

Choose $0<\rho <1\mathrm{/}(32A)$ and a finite coordinate-grid cover of $S$ by boxes of radius $\rho$ with centers $a_i\in S$, $1\le i\le m$. In each coordinate form a left increasing binomial factor with threshold $a_{i,j}-3\rho \mathrm{/}2$ and a right decreasing factor with threshold $a_{i,j}+3\rho \mathrm{/}2$; omit the left factor when its threshold is at most zero and the right factor when its threshold is at least one. Their product $b_i$ has degree at most $4n$, lies in $[0,1]$, equals $1+O(e^{-cn})$ on the inner box of radius $\rho$ and its first derivatives there are $O(n{e}^{-cn})$; outside the outer box of radius $2\rho$ its value and first derivatives are $O(n{e}^{-cn})$. The omitted boundary factors require no suppression on a side where the outer box does not meet $S$.

Let $T_{i,s}$ be the total-degree-$s$ Taylor polynomial at $a_i$. When $a_i$ is on the boundary, take limits of interior centers; all derivatives extend continuously to $S$. On the intersection of the outer box and $S$, convexity keeps the Taylor segment inside the triangle, so [F1] and the multinomial formula give $\mathrm{\mid }f-T_{i,s}\mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }(f-T_{i,s})\mathrm{\mid }\le CM(s+1)(4A\rho {)}^s\le CM{2}^{-s}$ after enlarging $C$ for small $s$. On all of $S$, the coefficient formula for $T_{i,s}$ gives $\mathrm{\mid }{T}_{i,s}\mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }{T}_{i,s}\mathrm{\mid }\le CM(s+1{)}^2{D}^s$ for a fixed $D>1$ depending only on $A$.

Set $w_i=b_i{\prod }_{j<i}(1-b_j)$ and $w_0={\prod }_{j\le m}(1-b_j)$. Then ${\sum }_{i=0}^m{w}_i=1$ and $0\le w_i\le 1$. At every point of $S$ some inner box applies, so one factor $1-b_i$, together with its first derivatives, is exponentially small there; the product rule and $\mathrm{\mid }{b}_j^{\mathrm{\prime }}\mathrm{\mid }\le Cn$ give $\mathrm{\parallel }{w}_0{\mathrm{\parallel }}_{W^{1,\mathrm{\infty }}(S)}\le Cn{e}^{-cn}$. Outside the outer box for a fixed $i$, the same product rule gives $\mathrm{\mid }{w}_i\mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }{w}_i\mathrm{\mid }\le Cn{e}^{-cn}$.

Put $q={\sum }_{i=1}^m{w}_i{T}_{i,s}$. Since $f-q=w_{0}f+{\sum }_i{w}_i(f-T_{i,s})$, the near-box and far-box bounds of steps 3.1 and 3.2 give $\mathrm{\parallel }f-q{\mathrm{\parallel }}_{W^{1,\mathrm{\infty }}(S)}\le CM\{n{2}^{-s}+n(s+1{)}^2{D}^s{e}^{-cn}\}\mathrm{.}$ Choose fixed $\epsilon >0$ with $\epsilon \le 1$ and $\epsilon \log D<c\mathrm{/}2$, and let $s=\lfloor \epsilon n\rfloor$. Then the right side is at most $CM{e}^{-b_{0}n}$. Each $w_i$ has degree at most $4mn$, so $q$ has total degree at most $4mn+s\le (4m+1)n$. Choose $n=\lfloor p\mathrm{/}(4m+1)\rfloor$; for the finitely many $p$ with $n=0$, the zero polynomial and an enlarged constant handle the claim. This proves the exponential estimate for every $p\ge 1$.

On the interval let $P$ be the $L^2$ orthogonal projection of $f^{\mathrm{\prime }}$ onto ${\mathbb{P}}_{p-1}$ and put $J_{p}f(t)=f(0)+{\int }_0^{t}P(s)ds$. The positive-definite finite polynomial Gram matrix gives a unique projection; Pythagoras gives $\mathrm{\parallel }P{\mathrm{\parallel }}_2\le \mathrm{\parallel }{f}^{\mathrm{\prime }}{\mathrm{\parallel }}_2$ and $\mathrm{\parallel }{f}^{\mathrm{\prime }}-P{\mathrm{\parallel }}_2\le \mathrm{\parallel }{f}^{\mathrm{\prime }}-r{\mathrm{\parallel }}_2$ for every $r\in {\mathbb{P}}_{p-1}$. Orthogonality to constants gives ${\int }_0^{1}P={\int }_0^1{f}^{\mathrm{\prime }}$, and [F4] evaluates the latter as $f(1)-f(0)$. Thus $J_{p}f(1)=f(0)+{\int }_0^{1}P=f(1)$, while $J_{p}f(0)=f(0)$ by construction. The derivatives of $f^{\mathrm{\prime }}$ satisfy a factorial bound with fixed constants $M^{\mathrm{\prime }}=MA$ and $A^{\mathrm{\prime }}=2A$, since $(j+1)!\le 2^{j}j!$ for $j\ge 1$. For $p\ge 2$, apply step 4.1 to $f^{\mathrm{\prime }}$ with degree $p-1$; best approximation of the $L^2$ projection then gives $\mathrm{\parallel }{f}^{\mathrm{\prime }}-P{\mathrm{\parallel }}_2\le CM{e}^{-b(p-1)}$, which is at most $C^{\mathrm{\prime }}M{e}^{-b^{\mathrm{\prime }}p}$ after changing constants. For $p=1$, the derivative contraction and the bound $\mathrm{\parallel }{f}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le MA$ give the same asserted estimate after enlarging $C^{\mathrm{\prime }}$. Finally $f-J_{p}f={\int }_0^t(f^{\mathrm{\prime }}-P)$ and [F2] bounds its $L^2$ norm by $\mathrm{\parallel }{f}^{\mathrm{\prime }}-P{\mathrm{\parallel }}_2$. The derivative and value bounds prove the stated interval estimate.