To prove Theorem 1.1, we first give a lemma.
Lemma 2.1
Let \(f\in W_{\infty }^{n}\). Assume that \(\Delta := \{a\le x_{1}< x_{2}<\cdots <x_{r}\le b,\alpha _{i}\in \mathbb{N},n=\sum_{i=1}^{r}\alpha _{i}\}\) is a Hermite interpolation system. Then, the remainder \(R_{\Delta }(f,x):=f(x)-H_{\Delta }(f,x)\) for the Hermite interpolation polynomial based on Δ satisfies
$$ \bigl\vert R_{\Delta }(f,x) \bigr\vert = \bigl\vert f(x)-H_{\Delta }(f,x) \bigr\vert \le \frac{ \Vert f^{(n)} \Vert _{\infty }}{n!} \bigl\vert W_{\Delta }(x) \bigr\vert ,\quad x\in [a,b], $$
(2.1)
where \(W_{\Delta }\) is given by (1.4). In particular, if \(f\in C^{n}\), then
$$ R_{\Delta }(f,x)=f(x)-H_{\Delta }(f,x)= \frac{f^{(n)}(\xi )}{n!}W_{ \Delta }(x),\quad x\in [a,b], $$
(2.2)
for some \(\xi \in [-1,1]\) depending on x and Δ.
Proof
Since (2.1) is trivially satisfied if x coincides with one of the interpolation points \(x_{1},\ldots ,x_{r}\), we need be concerned only with the case where x does not coincide with one of the interpolation nodes. Keeping x fixed, consider \(g:[a,b]\to \mathbb{R}\) given by
$$ g(y):=R_{\Delta }(f,y)-W_{\Delta }(y) \frac{R_{\Delta }(f,x)}{W_{\Delta }(x)},\quad y\in [a,b]. $$
(2.3)
By the assumption on f we know \(g\in W_{\infty }^{n}\). From (1.3) and (2.3) we conclude that g has at least \(n+1\) zeros (counting multiplicity), namely single zero x and \(\alpha _{k}\) fold zeros \(x_{k},k=1,\ldots ,r\). Then, by Rolle’s theorem, the derivative \(g'\) has at least n zeros. Repeating the argument, by induction we deduce that the derivative \(g^{(n-1)}\) has at least two zeros in \([a,b]\), which we denote by \(z_{1}\) and \(z_{2}\) (\(z_{1}< z_{2}\)), respectively. Since \(g\in W_{\infty }^{n}\), then by the Newton–Leibniz formula we obtain
$$ 0=g^{(n-1)}(z_{2})-g^{(n-1)}(z_{1})= \int _{z_{1}}^{z_{2}}g^{(n)}(y)\,dy. $$
(2.4)
It is known that \(H_{\Delta }(f)\) is an algebraic polynomial of degree at most \(n-1\). Hence, we obtain
$$ \bigl(H_{\Delta }(f)\bigr)^{(n)}(y)=0. $$
(2.5)
By a direct computation we obtain
$$ (W_{\Delta })^{(n)}(y)=n!. $$
(2.6)
Substituting (2.5) and (2.6) into (2.4), we obtain
$$ 0= \int _{z_{1}}^{z_{2}} \biggl[f^{(n)}(y)-n! \frac{R_{\Delta }(f,x)}{W_{\Delta }(x)} \biggr]\,dy= \int _{z_{1}}^{z_{2}}f^{(n)}(y)\,dy-n!(z_{2}-z_{1}) \frac{R_{\Delta }(f,x)}{W_{\Delta }(x)}. $$
(2.7)
From (2.7) it follows that
$$ R_{\Delta }(f,x)= \frac{\int _{z_{1}}^{z_{2}}f^{(n)}(y)\,dy}{n!(z_{2}-z_{1})}W_{\Delta }(x). $$
(2.8)
Combining
$$ \biggl\vert \int _{z_{1}}^{z_{2}}f^{(n)}(y)\,dy \biggr\vert \le \int _{z_{1}}^{z_{2}} \bigl\vert f^{(n)}(y) \bigr\vert \,dy \le \int _{z_{1}}^{z_{2}} \bigl\Vert f^{(n)} \bigr\Vert _{\infty }\,dy= \bigl\Vert f^{(n)} \bigr\Vert _{\infty }(z_{2}-z_{1}) $$
with (2.8) we obtain (2.1). Besides, if \(f\in C^{n}\), then \(g^{(n)}\) has at least one zero ξ in \([a,b]\), i.e., \(g^{(n)}(\xi )=0\). Hence, by differentiating n times on two sides of (2.3) first, and then substituting (2.5) and (2.6) into the obtained relation, we obtain
$$ 0=f^{(n)}(\xi )-n!\frac{R_{\Delta }(f,x)}{W_{\Delta }(x)}. $$
(2.9)
From (2.9) we obtain (2.2). This completes the proof of Lemma 2.1. □
Lemma 2.2
Let \(1\le p< \infty \) and assume that \(\omega (x)> 0\) is continuous-integrable on \((-1,1)\). Then, there exists a unique \(W_{n,p,\omega }\in \mathcal{P}_{n} \) for all \(n\in \mathbb{N}\) such that
$$ \Vert W_{n,p,\omega } \Vert _{p,\omega }=E_{n,p,\omega }\quad \textit{and}\quad W_{n,p, \omega }(x)=x^{n}+c_{1}x^{n-1}+ \cdots +c_{n}, $$
where \(E_{n,p,\omega }\) is given by (1.5). Furthermore, \(W_{n,p,\omega }\) has exactly n zeros given by (1.7).
Proof
The proof of the problem on \([-1,1]\) can be found in [16]. In general, we can use the variable substitution \(x=\frac{a+b}{2}+\frac{b-a}{2}t\) to refer the problem on \([a,b]\) to this on \([-1,1]\). We omit the details. □
Proof of Theorem 1.1
We consider (1) first. Let \(\Delta _{n,\infty }\) be given by (1.12). Then, for any \(f\in W_{\infty }^{n}\), it follows from (2.1) that
$$ \begin{aligned}[b] &\bigl\vert f(x)-L_{\Delta _{n,\infty }}(f,x) \bigr\vert \le \frac{ \Vert f^{(n)} \Vert _{\infty }}{n!} \Biggl\vert \prod_{i=1}^{n} \biggl(x-\frac{a+b}{2}-\frac{b-a}{2}\cos \frac{(2i-1)\pi }{2n} \biggr) \Biggr\vert , \\ &\quad x\in [a,b]. \end{aligned} $$
(2.10)
Let \(x=\frac{a+b}{2}+\frac{b-a}{2}t\). Then, (2.10) becomes
$$\begin{aligned} \bigl\vert f(x)-L_{\Delta _{n,\infty }}(f,x) \bigr\vert \le & \frac{ \Vert f^{(n)} \Vert _{\infty }(b-a)^{n}}{n!2^{n}} \Biggl\vert \prod _{i=1}^{n} \biggl(t-\cos \frac{(2i-1)\pi }{2n} \biggr) \Biggr\vert \\ =&\frac{ \Vert f^{(n)} \Vert _{\infty }(b-a)^{n}}{n!2^{2n-1}} \bigl\vert T_{n}(t) \bigr\vert , \quad t \in [-1,1], \end{aligned}$$
(2.11)
where \(T_{n}\) is the nth Chebyshev polynomial of the first kind, i.e., \(T_{n}(t)=\cos (n\arccos t)\). Let \(f\in BW_{\infty }^{n}\). Then, we have \(\|f^{(n)}\|_{\infty }\le 1\). Combining this fact with \(\|T_{n}\|_{\infty }=1\) as well as (2.11), we obtain
$$ e\bigl(BW_{\infty }^{n},L_{\Delta _{n,\infty }},L_{\infty } \bigr)=\sup_{f\in BW_{\infty }^{n}} \bigl\Vert f-L_{\Delta _{n,\infty }}(f) \bigr\Vert _{\infty }\le \frac{(b-a)^{n}}{n!2^{2n-1}}. $$
(2.12)
From (1.2) and (2.12) we obtain the upper estimate.
Now, we consider the lower estimate. Let \(\Delta := \{a\le x_{1}< x_{2}<\cdots <x_{r}\le b,\alpha _{i}\in \mathbb{N},n=\sum_{i=1}^{r}\alpha _{i}\}\) be an arbitrary Hermite interpolation system of cardinality n in \([a,b]\). Consider the function \(g(x)=\frac{x^{n}}{n!}\). Then, from \(g^{(n)}(x)=1\) and (2.2) it follows that \(g\in W_{\infty }^{n}\) and
$$ g(x)-H_{\Delta }(g,x)=\frac{W_{\Delta }(x)}{n!},\quad x\in [a,b]. $$
(2.13)
Let \(x=\frac{a+b}{2}+\frac{b-a}{2}t\). Then, by (1.4) we obtain
$$ W_{\Delta }(x)=x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+ \cdots +a_{n}= \frac{(b-a)^{n}}{2^{n}}h(t),\quad t\in [-1,1],$$
(2.14)
where
$$ h(t)=t^{n}+b_{1}t^{n-1}+b_{2}t^{n-2}+ \cdots +b_{n}. $$
(2.15)
Then, it follows from Theorem 6.1 in [4, Ch. 3] that
$$ \Vert h \Vert _{\infty }\ge 2^{1-n}.$$
(2.16)
Combining (1.1), (2.13), (2.14) and (2.16), we obtain
$$ e\bigl(BW_{\infty }^{n}, H_{\Delta }, L_{\infty }\bigr)\ge \bigl\Vert g-H_{\Delta }(g) \bigr\Vert _{\infty }=\frac{ \Vert W_{\Delta } \Vert _{\infty }}{n!}=\frac{(b-a)^{n}}{n!2^{n}} \Vert h \Vert _{\infty }\ge \frac{(b-a)^{n}}{n!2^{2n-1}}.$$
(2.17)
From (1.2) and (2.17) we obtain the lower estimate.
Next, we consider (2). We consider the upper estimate first. Let \(\Delta _{n,p,\omega }\) be given by (1.8) and \(W_{n,p,\omega }\) be given by (1.6). If \(f\in BW_{\infty }^{n}\), then we have \(\|f^{(n)}\|_{\infty }\le 1\). Combining this fact with (2.1) we obtain
$$ \bigl\vert f(x)-L_{\Delta _{n,p,\omega }}(f,x) \bigr\vert \le \frac{ \vert W_{n,p,\omega }(x) \vert }{n!}, \quad x\in [a,b]. $$
It follows that
$$ \bigl\Vert f-L_{\Delta _{n,p,\omega }}(f) \bigr\Vert _{p,\omega }\le \frac{ \Vert W_{n,p,\omega } \Vert _{p,\omega }}{n!}= \frac{E_{n,p,\omega }}{n!}. $$
(2.18)
From (1.1) and (2.18) we obtain
$$ e\bigl(BW_{\infty }^{n},L_{\Delta _{n,p,\omega }},L_{p,\omega } \bigr)\le \frac{E_{n,p,\omega }}{n!}. $$
(2.19)
From (1.2) and (2.19) we obtain the upper estimate.
Now, we consider the lower estimate. Let \(\Delta := \{a\le x_{1}< x_{2}<\cdots <x_{r}\le b,\alpha _{i}\in \mathbb{N},n=\sum_{i=1}^{r}\alpha _{i}\}\) be an arbitrary Hermite interpolation system of cardinality n in \([a,b]\). Consider the function \(g(x)=\frac{x^{n}}{n!}\). Then, \(g\in W_{\infty }^{n}\) and (2.13) holds. From the first equality in (2.14) and (1.5) as well as (1.6) it follows that
$$ \Vert W_{\Delta } \Vert _{p,\omega }\ge E_{n,p,\omega }. $$
(2.20)
From (1.1), (2.13) and (2.20) it follows that
$$ e\bigl(BW_{\infty }^{n}, H_{\Delta }, L_{p,\omega }\bigr)\ge \bigl\Vert g-H_{\Delta }(g) \bigr\Vert _{p, \omega }=\frac{ \Vert W_{\Delta } \Vert _{p,\omega }}{n!}\ge \frac{E_{n,p,\omega }}{n!}.$$
(2.21)
From (1.2) and (2.21) we obtain the lower estimate of (2). Theorem 1.1 is proved. □
Let \(BC^{n}=\{f\in C^{n}:\|f^{(n)}\|_{\infty }\le 1\}\). Using the fact \(BW_{\infty }^{n}\subset BC^{n}\) and \(g(x)=\frac{x^{n}}{n!}\in BC^{n}\) for \(n\in \mathbb{N}\), combining the proof of Theorem 1.1, we obtained the following results.
Corollary 2.3
-
(1)
For \(p= \infty \), we have
$$ e\bigl(n,BC^{n} ,L_{\infty }\bigr)=e\bigl(BC^{n},L_{\Delta _{n,\infty }}, L_{\infty }\bigr)= \frac{(b-a)^{n}}{n!2^{2n-1}},$$
(2.22)
where \(\Delta _{n,\infty }\) is given by (1.12).
-
(2)
Let \(1\le p< \infty \) and assume that \(\omega (x)> 0\) is continuous-integrable on \((a,b)\). Then, we have
$$ e\bigl(n,BC^{n} , L_{p,\omega }\bigr)=e\bigl(BC^{n},L_{\Delta _{n,p,\omega }}, L_{p, \omega }\bigr)= \frac{E_{n,p,\omega }}{n!},$$
(2.23)
where \(\Delta _{n,p,\omega }\) is given by (1.8).
Proof of Theorem 1.2
We consider (1) first. For any \(f\in BW_{\infty }^{n}\), from (2.1) it follows that
$$ \begin{aligned}[b] &\bigl\vert f(x)-L_{\Delta ^{*}_{n,\infty }}(f,x) \bigr\vert \le \frac{1}{n!} \Biggl\vert \prod_{i=1}^{n} \biggl(x-\frac{a+b}{2}-\frac{b-a}{2}\cos \frac{(2i-1)\pi }{2n} \Big/ \cos \frac{\pi }{2n} \biggr) \Biggr\vert , \\ &\quad x\in [a,b]. \end{aligned} $$
(2.24)
Let \(x=\frac{a+b}{2}+\frac{b-a}{2\cos \frac{\pi }{2n}}t\). Then, we have
$$ \begin{aligned}[b] &\prod_{i=1}^{n} \biggl(x-\frac{a+b}{2}-\frac{b-a}{2}\cos \frac{(2i-1)\pi }{2n} \Big/ \cos \frac{\pi }{2n} \biggr)= \frac{(b-a)^{n}T_{n}(t)}{ (\cos \frac{\pi }{2n} )^{n}2^{2n-1}}, \\ &\quad t\in \biggl[-\cos \frac{\pi }{2n},\cos \frac{\pi }{2n} \biggr]. \end{aligned} $$
(2.25)
From (1.1), (2.24) and (2.25) it follows that
$$ \begin{aligned}[b] e\bigl(BW_{\infty }^{n}, L_{\Delta ^{*}_{n,\infty }}, L_{\infty }\bigr)&\le \frac{(b-a)^{n}}{ (\cos \frac{\pi }{2n} )^{n}2^{2n-1}n!} \sup_{t\in [-\cos \frac{\pi }{2n},\cos \frac{\pi }{2n} ]} \bigl\vert T_{n}(t) \bigr\vert \\ & = \frac{(b-a)^{n}}{ (\cos \frac{\pi }{2n} )^{n}2^{2n-1}n!}. \end{aligned} $$
(2.26)
From (1.2) and (2.26) we obtain the upper estimate.
Now, we consider the lower estimate. Let \(\Delta := \{a= x_{1}< x_{2}<\cdots <x_{r}= b,\alpha _{i}\in \mathbb{N},n=\sum_{i=1}^{r}\alpha _{i}\}\) be an arbitrary Hermite interpolation system of cardinality n including the endpoints a and b. Consider the function \(g(x)=\frac{x^{n}}{n!}\). Then, \(g\in W_{\infty }^{n}\) and (2.13) holds. Let \(x=\frac{a+b}{2}+\frac{b-a}{2}t\). Denote \(t_{i}=\frac{2}{b-a} (x_{i}-\frac{a+b}{2} )\), \(i=1,\ldots ,r\). Then, by (1.4) one obtains
$$ W_{\Delta }(x)=\frac{(b-a)^{n}}{2^{n}}\prod _{i=1}^{r}(t-t_{i})^{ \alpha _{i}}, \quad t_{1}=-1,\, t_{r}=1,\, t\in [-1,1].$$
(2.27)
Let
$$ g(t)=\bigl(t^{2}-1\bigr)\prod_{i=2}^{n-1} \biggl(t-\cos \frac{(2i-1)\pi }{2n} \Big/\cos \frac{\pi }{2n} \biggr)= \frac{T_{n}(t\cos \frac{\pi }{2n})}{2^{n-1} (\cos \frac{\pi }{2n} )^{n}}. $$
Then, it is easy to verify that
$$ \Vert g \Vert _{\infty }= \frac{1}{2^{n-1} (\cos \frac{\pi }{2n} )^{n}} $$
(2.28)
and
$$ g \biggl(\frac{\cos \frac{i\pi }{n}}{\cos \frac{\pi }{2n}} \biggr)= \frac{(-1)^{i}}{2^{n-1} (\cos \frac{\pi }{2n} )^{n}},\quad i=1, \ldots ,n-1. $$
(2.29)
Assume that
$$ \Biggl\Vert \prod_{i=1}^{r}(t-t_{i})^{\alpha _{i}} \Biggr\Vert _{\infty }< \frac{1}{2^{n-1} (\cos \frac{\pi }{2n} )^{n}}. $$
(2.30)
Let
$$ R(t)=g(t)-\prod_{i=1}^{r}(t-t_{i})^{\alpha _{i}}, \quad t\in [-1,1]. $$
Then, it is easy to verify that \(R(t)\) is a polynomial of degree at most \(n-1\). Furthermore, from (2.29) and (2.30) one can check that
$$ R \biggl(\frac{\cos \frac{i\pi }{n}}{\cos \frac{\pi }{2n}} \biggr) (-1)^{i}>0,\quad i=1, \ldots ,n-1. $$
Thus, the polynomial \(R(t)\) has at least \(n-2\) zeros in \((-1,1)\). As \(t_{1}=-1\), \(t_{r}=1\), it is clear that ±1 are zeros of \(R(t)\). Hence, \(R(t)\) has at least n zeros in \([-1,1]\). This, and the fact that \(R(t)\) is a polynomial of degree at most \(n-1\), implies that \(R(t)=0\). Therefore,
$$ \Biggl\Vert \prod_{i=1}^{r}(t-t_{i})^{\alpha _{i}} \Biggr\Vert _{\infty }= \Vert g \Vert _{\infty }= \frac{1}{2^{n-1} (\cos \frac{\pi }{2n} )^{n}}, $$
which contradicts (2.30). Hence, we have
$$ \Biggl\Vert \prod_{i=1}^{r}(t-t_{i})^{\alpha _{i}} \Biggr\Vert _{\infty }\ge \frac{1}{2^{n-1} (\cos \frac{\pi }{2n} )^{n}}. $$
(2.31)
From (1.1), (2.13), (2.27) and (2.31) we obtain
$$ e\bigl(BW_{\infty }^{n}, H_{\Delta }, L_{\infty }\bigr)\ge \bigl\Vert g-H_{\Delta }(g) \bigr\Vert _{\infty }=\frac{ \Vert W_{\Delta } \Vert _{\infty }}{n!}\ge \frac{(b-a)^{n}}{ (\cos \frac{\pi }{2n} )^{n}2^{2n-1}n!}.$$
(2.32)
From (1.2) and (2.32) we obtain the lower estimate of (1).
Next, we consider (2). Let ω̅ and \(\Delta ^{*}_{n,p,\omega }\) be given by (1.19). Then, for any \(f\in BW_{\infty }^{n}\), from (2.1) it follows that
$$ \bigl\vert f(x)-L_{\Delta ^{*}_{n,p,\omega }}(f,x) \bigr\vert \le \frac{(1-x^{2}) \vert W_{n-2,p,\overline{\omega }}(x) \vert }{n!},\quad x\in [a,b]. $$
(2.33)
From (2.33) it follows that
$$ \bigl\Vert f-L_{\Delta ^{*}_{n,p,\omega }}(f) \bigr\Vert _{p,\omega }\le \frac{ \Vert W_{n-2,p,\overline{\omega }} \Vert _{p,\overline{\omega }}}{n!}= \frac{E_{n-2,p,\overline{\omega }}}{n!}. $$
(2.34)
From (1.1) and (2.34) we conclude that
$$ e\bigl(BW_{\infty }^{n},L_{\Delta ^{*}_{n,p,\omega }},L_{p,\omega } \bigr)\le \frac{E_{n-2,p,\overline{\omega }}}{n!}. $$
(2.35)
On the other hand, let \(\Delta := \{a= x_{1}< x_{2}<\cdots <x_{r}= b,\alpha _{i}\in \mathbb{N},n=\sum_{i=1}^{r}\alpha _{i}\}\) be an arbitrary Hermite interpolation system of cardinality n including the endpoints. Consider the function \(g(x)=\frac{x^{n}}{n!}\). Then, \(g\in W_{\infty }^{n}\) and (2.13) holds. From (1.1), (2.13), (1.5) and (1.6) it follows that
$$\begin{aligned} e\bigl(BW_{\infty }^{n}, H_{\Delta }, L_{p,\omega }\bigr) \ge & \bigl\Vert g-H_{\Delta }(g) \bigr\Vert _{p, \omega }= \frac{1}{n!} \Biggl\Vert \prod_{k=1}^{r}(x-x_{k})^{\alpha _{k}} \Biggr\Vert _{p,\omega } \\ =& \frac{1}{n!} \Biggl\Vert (x-a)^{\alpha _{1}-1}(b-x)^{\alpha _{r}-1} \prod_{k=2}^{r-1}(x-x_{k})^{\alpha _{k}} \Biggr\Vert _{p,\overline{\omega }} \\ \ge & \frac{1}{n!} \Biggl\Vert \prod_{k=1}^{n-2}(x- \xi _{k,p, \overline{\omega }}) \Biggr\Vert _{p,\overline{\omega }}= \frac{E_{n-2,p,\overline{\omega }}}{n!}. \end{aligned}$$
(2.36)
From (2.35) and (2.36) as well as (1.2) we obtain the result of (2). Theorem 1.2 is proved. □
Using the fact that \(BC^{n}\subset BW_{\infty }^{n}\) and \(g(x)=\frac{x^{n}}{n!}\in BC^{n}\) for \(n\in \mathbb{N}\), combining the proof of Theorem 1.2, we obtained the following results.
Corollary 2.4
-
(1)
Let \(p= \infty \) and \(n>2\). Then, we have
$$ \overline{e}\bigl(n,BC^{n}, L_{\infty }\bigr) =e \bigl(BC^{n}, L_{{\Delta ^{*}_{n, \infty }}}, L_{\infty }\bigr)= \frac{(b-a)^{n}}{ (\cos \frac{\pi }{2n} )^{n}2^{2n-1}n!},$$
where \(\Delta ^{*}_{n,\infty }\) is given by (1.17).
-
(2)
Let \(1\le p< \infty \), \(n>2\) and assume that \(\omega (x)> 0\) is continuous-integrable on \((a,b)\). Then, we have
$$ \overline{e}\bigl(n,BC^{n}, L_{p,\omega }\bigr) =e \bigl(BC^{n}, L_{\Delta ^{*}_{n,p, \omega }}, L_{p,\omega }\bigr)= \frac{E_{n-2,p,\overline{\omega }}}{n!},$$
where ω̅ and \(\Delta ^{*}_{n,p,\omega }\) are given by (1.19).
Remark 2.5
When \(n\ne r\), the nth optimal Hermite interpolation system of the problems given by (1.2) and (1.15) for \(BW_{\infty }^{r}\) in \(L_{\infty }\) and \(L_{p,\omega }\) (\(1\le p<\infty \)) are open problems.
Remark 2.6
When \(n= r\), the nth optimal Birkhoff interpolation system of the problems given by (1.2) and (1.15) for \(BW_{\infty }^{n}\) in \(L_{\infty }\) and \(L_{p,\omega }\) (\(1\le p<\infty \)) are open problems.