inequalities – “Reversed” Bernstein Inequality

I’m studying harmonic analysis by myself, and I read some online notes that introduce the Bernstein inequality. One of them mention a reversed form of the Bernstein inequality, which is stated below:

Let $mathbb{T} = mathbb{R} / mathbb{Z} = (0,1)$ be the one-dimensional torus. Assume that a function $f in L^{1}(mathbb{T})$ satisfies $hat{f}(j) = 0$ for all $|j| < n$ (vanishing Fourier coefficients). Then for all $1 leq p leq infty$, there exists some constant $C$ independent of $n,p$ and $f$, such that
$$||f’||_{p} geq Cn||f||_{p}$$

It seems that an easier problem can be obtained by replacing $f’$ with $f”$ in the above inequality. The easier problem is addressed in the MO post below:

Does there exist some $C$ independent of $n$ and $f$ such that $ |f”|_p geq Cn^2 | f |_p$, where $1 leq pleq infty$?

However, it seems that the trick of convex Fourier coefficients used in the post above no longer applies to the harder problem (lower bounding the norm of the first derivative). Any suggestions/ideas?

fa.functional analysis – Why the sequence of Bernstein polynomials of $sqrt x$ is increasing?

Bernstein polynomials preserves nicely several global properties of the function to be approximated: if e.g. $f:(0,1)tomathbb R$ is non-negative, or monotone, or convex; or if it has, say, non-negative $17$-th derivative, on $(0,1)$, it is easy to see tat the same holds true for the polynomials $B_nf$. On the contrary, comparing $B_nf$ and $B_{n+1}f$ turns out to be harder, due to the different choice of nodes where $f$ is evaluated. Consider for instance the Bernstein polynomials of the function $sqrt{x}$,
$$p_n(x):=sum_{k=0}^n{nchoose k}sqrt{frac kn},x^k(1-x)^{n-k}.$$

Question: Is this sequence of polynomial increasing? More generally, when is $B_nf$ increasing?

Some tentative approaches and remarks.

1. To compare $p_{n+1}$ with $p_n$ we may write the binomial coefficients in the expression for $p_{n+1}(x)$ as ${n+1choose k}={nchoose k}+{nchoose k-1}$; splitting correspondingly the sum into two sums, and shifting the index in the latter,
we finally get
$$p_{n+1}(x)-p_n(x)=sum_{k=0}^n{nchoose k}bigg(xsqrt{frac{k+1}{n+1}}+(1-x)sqrt{frac k{n+1}}-sqrt{frac kn},bigg)x^k(1-x)^{n-k},$$
which at least has non-negative terms approximatively for $frac kn<x$, which is still not decisive.

2. Monotonicity of the sequence $B_nf$ is somehow reminiscent of that of the real exponential sequence $big(1+frac xnbig)^n$. Precisely, let $delta_n:fmapsto frac{f(cdot+frac1n)-f(cdot)}{frac1n}$ denote the discrete difference operator, and $e_0:fmapsto f(0)$ the evaluation at $0$. Then the Bernstein operator $fmapsto B_nf$ can be written as $B_n=e_0displaystyle Big({bf 1} + frac{xdelta_n}nBig)^n$ (which, at least for analytic functions, converges to the Taylor series $e^{xD}$ at $0$). Unfortunately, the analogy seems to stop here.

3. The picture below shows the graphs of $sqrt x$ and of the first ten $p_n(x)$.
(The convergence is somehow slow; indeed it is $O(n^{-1/4})$, as per Kac’ general estimate $O(n^{-frac alpha2})$ for $alpha$-Hölder functions). The picture leaves some doubts about the endpoints; yet there should be no surprise, since $p_n(0)=0$, $p_n'(0)=sqrt{n}uparrow+infty$, $p_n(1)=1$, $p_n'(1)=frac1{1+sqrt{1-frac1n}}downarrowfrac12$.


python – Empirical probability density function and fitting the distribution with an approximating polynomial and via a Bernstein approach.?

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Approximation of a function by a polynomial (Chebyshev First Kind, Bernstein, etc…) containing only even degrees and constants in a given Range[a,b]

In Mathematica, how can I create a polynomial function containing only even degrees and constants?

That is, I have a function:

$f(x)=frac{pi ^2}{left(frac{pi }{2}-tan ^{-1}(k (x-1))right)^2}$

And I’m looking for a function that generates such a polynomial approximation on arbitrary range with Chebyshev First Kind, Bernstein or another in form:

$p(x) = c_0 + c_1 cdot x^2+ … + c_m cdot x^m$

where $m$ – maximum even degree of polynom.

pars = {k = 1, Subscript((Omega), 0) = 1, (CapitalDelta) = 5}
f = (-ArcTan(k (x - Subscript((Omega), 0))) + Pi/2)/Pi
Plot(f, {x, 0, 5}, PlotRange -> All)
p(x_) = 1/f^2
Plot(1/f^2, {x, 0, 5}, PlotRange -> All)
P = Collect(
   N(InterpolatingPolynomial({{0, 0}, {Subscript((Omega), 0)/2, 
       p(Subscript((Omega), 0)/2)}, {Subscript((Omega), 0), 
       p(Subscript((Omega), 0))}, {(CapitalDelta)/2, 
       p((CapitalDelta)/2)}, {(CapitalDelta), p((CapitalDelta))}}, 
     x)), x) // Simplify
Plot({1/f^2, P}, {x, 0, 2}, PlotRange -> All)

Somewhere on the Internet I read information that in order to get rid of odd degrees you need to include the point {0,0} in the polynomial, which I did with the usual InterpolatingPolynomial command.
Despite this, the odd degrees in the polynomial have been preserved.