# 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. $$f:[0,1]\to\mathbb R$$ 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 that the same holds true for the polynomials $$B_nf$$. In particular, since all $$B_n$$ fix all affine functions, if $$f\le ax+b$$, also $$B_nf(x)\le ax+b$$, whence it follows immediately $$B_nf\le f$$ for concave $$f$$. On the other hand, 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 $$\sqrt{x}$$, $$p_n(x):=\sum_{k=0}^n{n\choose k}\sqrt{\frac kn}\,x^k(1-x)^{n-k}.$$

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 $${n+1\choose k}={n\choose k}+{n\choose k-1}$$; splitting correspondingly the sum into two sums, and shifting the index in the latter, we finally get $$p_{n+1}(x)-p_n(x)=\sum_{k=0}^n{n\choose k}\bigg[x\sqrt{\frac{k+1}{n+1}}+(1-x)\sqrt{\frac k{n+1}}-\sqrt{\frac kn}\,\bigg]x^k(1-x)^{n-k},$$ 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 $$\big(1+\frac xn\big)^n$$. Precisely, let $$\delta_n:f\mapsto \frac{f(\cdot+\frac1n)-f(\cdot)}{\frac1n}$$ denote the discrete difference operator, and $$e_0:f\mapsto f(0)$$ the evaluation at $$0$$. Then the Bernstein operator $$f\mapsto B_nf$$ can be written as $$B_n=e_0\displaystyle \Big({\bf 1} + \frac{x\delta_n}n\Big)^n$$ (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 $$p_n(0)=0$$, $$p_n'(0)=\sqrt{n}\uparrow+\infty$$, $$p_n(1)=1$$, $$p_n'(1)=\frac1{1+\sqrt{1-\frac1n}}\downarrow\frac12$$.

• If $f$ is concave on $[0,1]$ then Bernstein polynomials $B_{n}(f,x)$ is increasing in $n$. Coincidentally I had this question on a homework (see problem 5 in math.uci.edu/~pivanisv/EX5W2019.pdf) I will try to post the answer later today. Jan 16, 2021 at 17:06
• We can write a definite integral formula for $p_n(x)$ starting from $\sqrt a = (2\sqrt\pi)^{-1} \int_0^\infty (1 - e^{-at}) \, dt/t^{3/2}$; one choice if I did this right is $p_n(x) = (2\sqrt\pi)^{-1} \{ 1 - [1 - (1-e^{-t/n}) x]^n \} \, dt/t^{3/2}$. This can be used to estimate $p_n(x)$, but it might not make it any easier to decide whether $p_{n+1}(x) > p_n(x)$ for all $x,n$. Jan 16, 2021 at 17:20
• @PietroMajer, This paper ijnaa.semnan.ac.ir/… gives references. They are referring to Temple (1954), I guess it is this one projecteuclid.org/euclid.dmj/1077465882 however I do not have access to it, not sure what is Temple's proof but I would be also curious to know. Jan 17, 2021 at 15:29
• @PietroMajer yes. Also just a remark that the argument I presented proves this stronger statement: in the converse implication I used only the fact that $B_{n} f \leq f$. Jan 17, 2021 at 17:49
• yes, also, you only need to assume the inequality for large $n$, if I understand correctly Jan 17, 2021 at 17:53

As noted by Paata Ivanishvili, if $$f$$ is concave on $$[0,1]$$, then the Bernstein polynomials $$B_n(f,p)$$ are increasing in $$n$$. Here is a probabilistic proof:

Let $$I_j$$ for $$j \ge 1$$ be independent variables taking value 1 with probability $$p$$ and $$0$$ with probability $$1-p$$. Then $$X_n:=\sum_{j=1}^n I_j$$ has a binomial Bin$$(n,p)$$ distribution and the Bernstein polynomial can be written as $$B_n(f,p)=E[f(X_n/n)]$$. Now for every $$j \in [1,n+1]$$, the random variable $$Y_j=Y_j(n)=X_{n+1}-I_j$$ also has a Bin$$(n,p)$$ distribution and $${X_{n+1}} = {\sum_{j=1}^{n+1} (Y_j/n)} \, .$$ For concave $$f$$, Jensen's inequality gives $$f \left(\frac{\sum_{j=1}^{n+1} (Y_j/n)}{n+1} \right) \ge \left(\frac{\sum_{j=1}^{n+1} f(Y_j/n)}{n+1} \right)$$ whence $$B_{n+1}(f,p)=E f \left(\frac{X_{n+1}}{n+1}\right)=E f \left(\frac{\sum_{j=1}^{n+1} (Y_j/n)}{n+1} \right) \ge E \left(\frac{\sum_{j=1}^{n+1} f(Y_j/n)}{n+1} \right) =B_n(f,p)$$

• I'm still staring at it... Jan 16, 2021 at 18:17
• Is there some part of the argument you did not find convincing? Jan 17, 2021 at 3:14
• I could not believe there was so a chrystal clear explanation... Thank you! Jan 17, 2021 at 6:51

While nothing will beat the brilliant probabilistic proof given in Yuval Peres's answer, a more conventional argument goes as follows. Write $$a_{n,k} = \tbinom nk x^k (1-x)^{n-k}$$ and $$p_{n,k} = \tfrac{k}{n}$$ Observe that $$\tfrac{k}{n} = (1 - p_{n,k}) \tfrac{k}{n-1} + p_{n,k} \tfrac{k-1}{n-1} .$$ Thus, if $$f$$ is concave, then $$f(\tfrac{k}{n}) \geqslant (1 - p_{n,k}) f(\tfrac{k}{n-1}) + p_{n,k} f(\tfrac{k-1}{n-1}) .$$ Multiply this by $$a_{n,k}$$ and add up to get \begin{aligned} p_n(x) & = \sum_{k=0}^n a_{n,k} f(\tfrac{k}{n}) \\ & \geqslant \sum_{k=0}^n a_{n,k} \Bigl( (1 - p_{n,k}) f(\tfrac{k}{n-1}) + p_{n,k} f(\tfrac{k-1}{n-1}) \Bigr) \\ & = \sum_{k=0}^{n-1} a_{n,k} (1 - p_{n,k}) f(\tfrac{k}{n-1}) + \sum_{k=1}^n a_{n,k} p_{n,k} f(\tfrac{k-1}{n-1}) \\ & = \sum_{k=0}^{n-1} \bigl(a_{n,k} (1 - p_{n,k}) + a_{n,k+1} p_{n,k+1} \bigr) f(\tfrac{k}{n-1}) . \end{aligned} Now an elementary calculation reduces the right-hand side to $$p_{n-1}(x)$$.

(I make this answer CW, since I expect this is exactly the solution Paata had in mind.)

• Yes, that is the solution I had in mind. Jan 16, 2021 at 22:46
• @mathworker21: Ouch! Thanks for spotting this. Jan 16, 2021 at 22:48
• @PaataIvanishvili: In that case, and if you care about points, feel free to post your answer, and I'll delete this one. :-) Jan 16, 2021 at 22:49
• Thanks, no need to delete. I do not need points. Perhaps I will post tho converse implication that Pietro asked in the comments Jan 16, 2021 at 22:51
• This is also what I was aiming to in the past days! Unfortunately, due to a wrong numerical experiment, I falsely convinced myself that the result was not generally true for all convex functions, but needed additional assumptions. Sometimes one is close to the truth and doesn't see it :'( Lucky there is MO! Jan 17, 2021 at 0:35

I would also be happy to know the converse implication, that you quoted in the last remark of your notes

Let $$f \in C([0,1])$$. Then $$p_{n+1}(f,x) \geq p_{n}(f,x)$$ for all $$n \in \mathbb{N}$$ and all $$x \in [0,1]$$ if and only if $$f$$ is concave. Yuval gave a nice proof of one implication. To show the converse, assume contrary that $$f$$ is not concave. By adding a linear function if necessary, and scaling the result by a positive constant we can assume that there exists $$x \in [a',b'] \subseteq [0,1]$$ such that $$f(a')=f(b')>1$$, and

$$0=\min_{[a',b']}f = f(x) < \frac{x-a'}{b'-a'}f(b')+\frac{b'-x}{b'-a'}f(a')$$. By continuity we can further find $$a such that $$0\leq a' so that $$\min\limits _{[a'b']\setminus [a,b]} f \geq 1$$. Then, I claim that for $$n$$ large enough we must have $$p_{n}(f,x)>0$$. Indeed,

$$p_{n}(f,x) = \sum_{0\leq k \leq n} f\left(\frac{k}{n}\right)\binom{n}{k}x^{k}(1-x)^{n-k} \geq -\|f\|_{C}\, n \max\limits_{0\leq k \leq [a'n]}\binom{n}{k}x^{k}(1-x)^{n-k}+\binom{n}{[an]}x^{[an]}(1-x)^{n-[an]}+ \binom{n}{[bn]}x^{[bn]}(1-x)^{n-[bn]} -\|f\|_{C}\, n\max\limits_{[b'n]\leq k \leq n}\binom{n}{k}x^{k}(1-x)^{n-k}$$ Let us show that the sum of the first two terms in the right hand side of the inequality is positive (for $$n$$ large enough), similarly the sum of the last two terms will be positive. Indeed, to verify

$$\binom{n}{[an]}x^{[an]}(1-x)^{n-[an]} >\|f\|_{C}\, n \max\limits_{0\leq k \leq [a'n]}\binom{n}{k}x^{k}(1-x)^{n-k}$$ Take $$1/n$$ power and let $$n \to \infty$$. Since $$\binom{n}{sn}^{1/n} \to s^{-s}(1-s)^{s-1}$$ as $$n \to \infty$$ it suffices to verify that

$$\max_{0\leq s \leq a'}s^{-s}(1-s)^{s-1}x^{s} (1-x)^{1-s} < a^{-a}(1-a)^{a-1}x^{a}(1-x)^{1-a}$$ Notice that $$\varphi(s) := s^{-s}(1-s)^{s-1}x^{s} (1-x)^{1-s}$$ is increasing on $$[0,a]$$. Indeed,

$$(\ln \varphi (s) )' = \ln \left(\frac{1-s}{s} \cdot \frac{x}{1-x}\right) >0,$$ where the last inequality follows because $$x>a\geq s$$.

Thus for $$n$$ large enough $$p_{n}(f,x)>0$$. On the other hand, as Yuval pointed out, $$p_{n}(f,x) = \mathbb{E}f(\frac{\xi_{1}+...+\xi_{n}}{n}) \to \mathbb{E}f(\mathbb{E} \xi_{1})=f(x)=0$$ by the strong law of large numbers. Since $$n \mapsto p_{n}(f,x)$$ increasing we have $$p_{n}(f,x) \leq 0$$ which is in contradiction with $$p_{n}(f,x)>0$$ for $$n$$ large enough.

• This is also very interesting. Thank you! Jan 17, 2021 at 8:50