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, 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.

``````ClearAll("Global`*")
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.