Whitney inequality

Denote the value of the best uniform approximation of a function ${\displaystyle f\in C([a,b])}$ by algebraic polynomials ${\displaystyle P_{n}}$ of degree ${\displaystyle \leq n}$ by

${\displaystyle E_{n}(f)_{[a,b]}:=\inf _{P_{n}}{\|f-P_{n}\|_{C([a,b])}}}$

The moduli of smoothness of order ${\displaystyle k}$ of a function ${\displaystyle f\in C([a,b])}$ are defined as:

${\displaystyle \omega _{k}(t):=\omega _{k}(t;f;[a,b]):=\sup _{h\in [0,t]}\|\Delta _{h}^{k}(f;\cdot )\|_{C([a,b-kh])}\quad {\text{ for }}\quad t\in [0,(b-a)/k],}$

${\displaystyle \omega _{k}(t):=\omega _{k}((b-a)/k)\quad {\text{ for}}\quad t>(b-a)/k,}$

where ${\displaystyle \Delta _{h}^{k}}$ is the finite difference of order ${\displaystyle k}$.

Theorem: [2] [Whitney, 1957] If ${\displaystyle f\in C([a,b])}$, then

${\displaystyle E_{k-1}(f)_{[a,b]}\leq W_{k}\omega _{k}\left({\frac {b-a}{k}};f;[a,b]\right)}$

where ${\displaystyle W_{k}}$ is a constant depending only on ${\displaystyle k}$. The Whitney constant ${\displaystyle W(k)}$ is the smallest value of ${\displaystyle W_{k}}$ for which the above inequality holds. The theorem is particularly useful when applied on intervals of small length, leading to good estimates on the error of spline approximation.

The original proof given by Whitney follows an analytic argument which utilizes the properties of moduli of smoothness. However, it can also be proved in a much shorter way using Peetre's K-functionals.[3]

Let:

${\displaystyle x_{0}:=a,\quad h:={\frac {b-a}{k}},\quad x_{j}:=x+0+jh,\quad F(x)=\int _{a}^{x}f(u)\,du,}$

${\displaystyle G(x):=F(x)-L(x;F;x_{0},\ldots ,x_{k}),\quad g(x):=G'(x),}$

${\displaystyle \omega _{k}(t):=\omega _{k}(t;f;[a,b])\equiv \omega _{k}(t;g;[a,b])}$

where ${\displaystyle L(x;F;x_{0},\ldots ,x_{k})}$ is the Lagrange polynomial for ${\displaystyle F}$ at the nodes ${\displaystyle \{x_{0},\ldots ,x_{k}\}}$.

Now fix some ${\displaystyle x\in [a,b]}$ and choose ${\displaystyle \delta }$ for which ${\displaystyle (x+k\delta )\in [a,b]}$. Then:

${\displaystyle \int _{0}^{1}\Delta _{t\delta }^{k}(g;x)\,dt=(-1)^{k}g(x)+\sum _{j=1}^{k}(-1)^{k-j}{\binom {k}{j}}\int _{0}^{1}g(x+jt\delta )\,dt}$

${\displaystyle =(-1)^{k}g(x)+\sum _{j=1}^{k}{(-1)^{k-j}{\binom {k}{j}}{\frac {1}{j\delta }}(G(x+j\delta )-G(x))},}$

Therefore:

${\displaystyle |g(x)|\leq \int _{0}^{1}|\Delta _{t\delta }^{k}(g;x)|\,dt+{\frac {2}{|\delta |}}\|G\|_{C([a,b])}\sum _{j=1}^{k}{\binom {k}{j}}{\frac {1}{j}}\leq \omega _{k}(|\delta |)+{\frac {1}{|\delta |}}2^{k+1}\|G\|_{C([a,b])}}$

And since we have ${\displaystyle \|G\|_{C([a,b])}\leq h\omega _{k}(h)}$, (a property of moduli of smoothness)

${\displaystyle E_{k-1}(f)_{[a,b]}\leq \|g\|_{C([a,b])}\leq \omega _{k}(|\delta |)+{\frac {1}{|\delta |}}h2^{k+1}\omega _{k}(h).}$

Since ${\displaystyle \delta }$ can always be chosen in such a way that ${\displaystyle h\geq |\delta |\geq h/2}$, this completes the proof.

It is important to have sharp estimates of the Whitney constants. It is easily shown that ${\displaystyle W(1)=1/2}$, and it was first proved by Burkill (1952) that ${\displaystyle W(2)\leq 1}$, who conjectured that ${\displaystyle W(k)\leq 1}$ for all ${\displaystyle k}$. Whitney was also able to prove that [2]

${\displaystyle W(2)={\frac {1}{2}},\quad {\frac {8}{15}}\leq W(3)\leq 0.7\quad W(4)\leq 3.3\quad W(5)\leq 10.4}$

and

${\displaystyle W(k)\geq {\frac {1}{2}},\quad k\in \mathbb {N} }$

In 1964, Brudnyi was able to obtain the estimate ${\displaystyle W(k)=O(k^{2k})}$, and in 1982, Sendov proved that ${\displaystyle W(k)\leq (k+1)k^{k}}$. Then, in 1985, Ivanov and Takev proved that ${\displaystyle W(k)=O(k\ln k)}$, and Binev proved that ${\displaystyle W(k)=O(k)}$. Sendov conjectured that ${\displaystyle W(k)\leq 1}$ for all ${\displaystyle k}$, and in 1985 was able to prove that the Whitney constants are bounded above by an absolute constant, that is, ${\displaystyle W(k)\leq 6}$ for all ${\displaystyle k}$. Kryakin, Gilewicz, and Shevchuk (2002)[4] were able to show that ${\displaystyle W(k)\leq 2}$ for ${\displaystyle k\leq 82000}$, and that ${\displaystyle W(k)\leq 2+{\frac {1}{e^{2}}}}$ for all ${\displaystyle k}$.