Home > Analysis, Complex analysis, Inequalities > Agregation 2013 – Analysis – Part 3

## Agregation 2013 – Analysis – Part 3

Part III: Muntz spaces and the Clarkson-Edros Theorem

Recall that for every ${\lambda \in \Bbb{N}}$ we define ${nu_\lambda(t)=t^\lambda,\ t \in [0,1]}$ and that ${\Lambda=(\lambda_n)_{n \in \Bbb{N}}}$ is a strictly increasing sequence of positive integers.

1. Suppose that ${\lambda_0=0}$ and ${\displaystyle \sum_{n \geq 1}\frac{1}{\lambda_n} =\infty}$. Let ${k \in \Bbb{N}\setminus \Lambda}$. Define ${Q_0=\nu_k}$ and by recurrence for ${n \in \Bbb{N}}$ define ${Q_{n+1}}$ by

$\displaystyle Q_{n+1}(x)=(\lambda_{n+1}x^{\lambda_{n+1}}\int_x^1 Q_n(t)t^{-1-\lambda_{n+1}}dt.$

(a) Calculate ${Q_1}$ and prove that ${\|Q_1\|_\infty \leq \displaystyle \left|1-\frac{k}{\lambda_1}\right|.}$

(b) Prove that for every ${n \geq 1}$ ${Q_n-\nu_k}$ is a linear combination of ${\nu_{\lambda_1},\nu_{\lambda_2},..,\nu_{\lambda_n}}$.

(c) Prove that for every ${n \geq 1}$ we have ${\displaystyle\|Q_n\|_\infty \leq \prod_{i=1}^n \left|1-\frac{k}{\lambda_j}\right|}$.

(d) Deduce that ${\nu_k \in \overline{M_\Lambda}}$.

(e) Conclude that ${C([0,1])=\overline{M_\Lambda}}$.

From here on suppose that ${\lambda_0}$ is arbitrary and the series ${\displaystyle \sum_{n \geq 1}\frac{1}{\lambda_n}}$ converges. For ${p \in \Bbb{N}}$ denote ${\rho_p(\Lambda)=\sum_{\lambda_n>p} \frac{1}{\lambda_n}}$. For ${s \in \Bbb{N}}$ denote ${N_s(\Lambda)}$ the cardinal of the set ${\{n \in \Bbb{N} | \lambda_n\leq s\}}$.

2. (a) Justify that ${\displaystyle \lim_{n \rightarrow \infty} \frac{N_s(\Lambda)}{s}=0}$.

(b) Compute ${\displaystyle \lim_{p \rightarrow \infty} \rho_p(\Lambda)}$.

3. Using the results in the second part prove that there exists a function ${\mu_\Lambda}$ continuous and bounded by ${1}$ on ${[0,1]}$ such that for every ${m \in \Bbb{N}}$ we have

$\displaystyle \int_0^1 x^m \mu_\Lambda(x)dx =\frac{2}{(m+2)^2}\prod_{n=0}^\infty \frac{\lambda_n-m}{\lambda_n+m+2}.$

4. (a) Prove that for every ${P \in M_\Lambda}$ we have ${\displaystyle \int_0^1 P(x)\mu_\Lambda(x)dx=0}$.

(b) Deduce that for every ${q \in \Bbb{N} \setminus \Lambda}$ we have

$\displaystyle \inf\{ \|\nu_q-P\|_\infty : P \in M_\Lambda\} \geq \left| \int_0^1 x^q \mu_\Lambda(x)dx \right|.$

5. For ${m}$ fixed prove that:

(a) ${\displaystyle \sum_{\lambda_n >2m} \ln \left( \frac{\lambda_n+m+2}{\lambda_n-m}\right) \leq 4(m+1)\rho_{2m}(\Lambda)}$.
Define ${\theta(x)=x\ln \displaystyle \left( \frac{3e}{x}\right)}$ for ${x>0}$ and ${\theta(0)=0}$ (by continuity).

(b) Prove that for ${k \geq 1}$ we have ${k! \geq k^ke^{-k}}$ and deduce that

$\displaystyle \prod_{m<\lambda_n \leq 2m} \frac{\lambda_n-m}{\lambda_n+m+2} \geq \frac{\tilde N!}{(3m+2)^{\tilde N}}\geq \exp(-(m+1)\theta(\tilde N/(m+1)))$

where ${\tilde N=N_{2m}(\Lambda)-N_m(\Lambda)}$.

For ${m \geq 1}$ we pose ${\hat N=N_{m-1}(\Lambda)}$ and we admit that we have in the same way the minoration

$\displaystyle \prod_{\lambda_n < m} \frac{\lambda_n-m}{\lambda_n+m+2} \geq \exp(-(m+1)\theta(\hat N/(m+1))).$

6. (a) Prove that there exists a sequence ${\gamma_m \rightarrow 0}$ such that for every ${m \in \Bbb{N}}$ we have

$\displaystyle \inf\{ \|\nu_m-P\|_{\infty} : P \in M_{\Lambda\setminus\{m\}}\} \geq e^{-(m+1)\gamma_m}.$

(b) Deduce from there that for ${m \in N\setminus \Lambda}$ we have ${\nu_m \notin \overline{M_\Lambda}}$.

7. Take ${\varepsilon >0}$ and prove that there exists ${\beta_\varepsilon>0}$ (depending only on ${\varepsilon}$ and ${\Lambda}$) such that for every ${P \in M_{\Lambda}}$ and for every ${k \in \Bbb{N}}$ we have

$\displaystyle |\alpha_k| \leq \beta_\varepsilon(1+\varepsilon)^{\lambda_k}\|P\|_\infty$

where ${P=\sum_{i \in \Bbb{N}} a_i\nu_{\lambda_i}}$ (with all ${a_i}$ zero from a point on). (hint: apply 6. (a))

Consider ${M\in \overline{M_\Lambda}}$ the limit of a sequence ${(P_n) \subset M_\Lambda}$ where for every ${n}$ the polynomial ${P_n}$ is written ${P_n =\sum_{ i \in \Bbb{N}} a_{n,i}\nu_{\lambda_i}}$ (where the coefficients of every polynomial are zero from a point on).

(b) Prove that for ${i}$ fixed, the sequence ${(a_{n,i})_n}$ converges to a complex number ${a_i}$.

(c) Deduce that for every ${x \in [0,1)}$ we have ${f(x)=\displaystyle \sum_{i=0}^\infty a_i x^{\lambda_i}}$.

8. Clarkson-Erdos Theorem

(a) Let ${f \in C([0,1])}$. For ${r \in [0,1)}$ we define ${f_r\in C([0,1])}$ by ${f_r(t)=f(rt)}$. Prove that ${\lim_{r \rightarrow 1}\|f_r-f\|_\infty=0}$.

(b) Prove that ${\overline{M_\Lambda}}$ is exactly the space of functions ${f}$ which are continuous on ${[0,1]}$ for which there exists a sequence of complex numbers ${(a_i)_i}$ such that

$\displaystyle f(x) = \sum_{i=0}^\infty a_ix^{\lambda_i},\ \text{ on }[0,1).$

Hints: (a) Simple integration yields ${Q_1(x)=x^k-x^{\lambda_1}}$. The estimate comes from

$\displaystyle |Q_1(x)| \leq |\lambda_{n+1}-k| \int_x^1 (t/x)^{-1-\lambda_{n+1}}\leq \frac{|\lambda_1-k|}{\lambda_1}.$

(we have used ${\|Q_1\|_\infty \leq 1}$.

(b) Simple induction.

(c) Use induction: (a) is the first step. To prove the induction step, use the recurrence relation and majorize ${Q_n}$ directly with the RHS just like we did in (a).

(d) For ${j}$ great enough we have ${1-k/\lambda_j >0}$ and moreover

$\displaystyle 1-k/\lambda_j \leq e^{-k/\lambda_j}.$

Therefore ${\|Q_n\|_\infty}$ is bounded by ${C\exp(\sum_{n \geq n_0} -k/\lambda_j)}$ which tends to zero, as the series ${\sum_j 1/\lambda_j}$ is divergent.

As a consequence there exists a sequence ${P_n \subset M_\Lambda}$ given by (b) ${\nu_k+P_n=Q_n}$ and the above estimate says that ${(-P_n)}$ converges uniformly to ${\nu_k}$ so ${\nu_k \in \overline{M_\Lambda}}$.

(e) From (d) and ${\nu_0 \in M_\Lambda}$ we conclude that all monomials are in ${M_\Lambda}$. (note that here we have used ${\lambda_0=0}$, and it is not possible to prove that ${\nu_0 \in M_\Lambda}$ using (c),(d), because in (c) the estimate doesn’t work for ${k=0}$) Hence, all polynomials are in ${M_\Lambda}$ and since the space of polynomials is dense in ${C([0,1])}$ for the uniform convergence, we are done.

2. (a) ${N_s(\Lambda)}$ is the cardinal of the set ${\{\frac{1}{s}\leq \frac{1}{\lambda_n}\}}$. From Part I.2 we deduce the desired result.

(b)

$\displaystyle \rho_p(\Lambda)=\sum_{n \in \Bbb{N}}\frac{1}{\lambda_n}-\sum_{\lambda_n < p}\frac{1}{\lambda_n}$

and since ${\lambda_n}$ is increasing, as ${p\rightarrow \infty}$ the sequence above has the limit equal to ${0}$.

3. As indicated in the official hint, we can use the second part, namely, the definition of ${F}$ and the final point II.6. We choose ${\alpha_n =1/(1+\lambda_n)}$ so that the series ${\sum\alpha_n}$ converges. Choose ${z=m+1}$ for which ${\text{Re}(z) >0}$, and we are done.

4. (a) If ${P=\nu_{\lambda_i}}$ then using the above formula we get

$\displaystyle \int_0^1 P(x)\mu_\Lambda(x)dx=0.$

this extends to every finite linear combination of monomials, and therefore to all ${M_\Lambda}$.

(b) One definition of ${\|\cdot \|_\infty}$ is the following:

$\displaystyle \|\nu_q-P\|_\infty = \sup_{\|u\|_{L^1}\leq 1} \int_0^1 (x^q-P)u(x)dx$

and since ${\|\mu_\Lambda\|_{L^1(0,1)}\leq 1}$ it is clear that

$\displaystyle \|\nu_q-P\|_\infty \geq \int_0^1 x^q \mu_\Lambda(x)dx,$

for every ${q \in \Bbb{N}\setminus \Lambda}$. (here we have used (a))

5. (a) Estimate term by term:

$\displaystyle \ln(\lambda_n+m+2)-\ln(\lambda_n-m) = \frac{1}{c}2(m+1),$

where ${\lambda_n-m\leq c}$. This means that

$\displaystyle \frac{1}{c}\leq \frac{1}{\lambda_n-m} < \frac{2}{\lambda_n},$

where the last inequality comes from the fact that ${\lambda_n > 2m}$. Summing up all these relations for ${\lambda_n>2m}$ we get the desired result.

(b) Use the inequality ${\displaystyle \left(\frac{k+1}{k}\right)^k and induction to prove that ${k! \geq k^k e^{-k}}$.

Since ${\lambda_n}$ is strictly increasing, if we denote by ${\lambda_{n_0+1},...\lambda_{n_0+\tilde N}}$ all the terms in the interval ${(m,2m]}$ then we have ${\lambda_{n_0+k}-m \geq k}$ so the product of all the denominators is greater than ${\widetilde N!}$.

All the denominators are smaller than ${3m+2}$ so the first part of the inequality is proved.

In order to prove the second part, let’s estimate

$\displaystyle -(m+1)\theta(\widetilde N/(m+1))=-\widetilde N \ln \frac{3e(m+1)}{\widetilde N}$

and then

$\displaystyle \exp(-(m+1)\theta(\widetilde N/(m+1))) = \left(\frac{3e(m+1)}{\widetilde N}\right)^{-\widetilde N}=e^{-\widetilde N} \frac{\widetilde N^{\widetilde N}}{(3m+3)^{\widetilde N}}.$

Now it should be clear what to do… Use the first inequality provided to see that

$\displaystyle \frac{\widetilde N !}{(3m+2)^{\widetilde N}} \geq \frac{\widetilde N^{\widetilde N}e^{-\widetilde N}}{(3m+2)^{\widetilde N}}.$

Now increase a bit the denominator and we are done. (do some similar reasoning for the part ${\lambda_n)

6. (a) Use 3,4 and 5 (formula for the infimum, formula for the integral, split product in 3 parts ${\lambda_n2m}$, and the corresponding estimates)to prove that a choice of gamma is

$\displaystyle \gamma_m= \theta(\widetilde N/(m+1))+\theta(\widehat N/(m+1))+4\rho_{2m}(\Lambda)-\frac{\ln(2/(m+1)^2)}{m+1}.$

(b) a clear consequence of (a).

7. (a) If we use 6. (a) as indicated for ${\nu_{\lambda_k}}$ and ${\displaystyle P=-\frac{1}{a_{k}}P}$ (note that if ${a_k=0}$ we have nothing to prove) we get

$\displaystyle \|P\|_\infty/|a_k| \geq e^{(-\lambda_k+1)\gamma_{\lambda_k}},$

and all that we have to do in order to prove the desired estimate is to prove that there exists ${\beta_\varepsilon>0}$ such that

$\displaystyle e^{(\lambda_k+1)\gamma_{\lambda_k}}\leq \beta_\varepsilon (1+\varepsilon)^{\lambda_k},$

which is equivalent to proving that

$\displaystyle \left(\frac{e^{\gamma_{\lambda_k} \frac{\lambda_k+1}{\lambda_k}}}{1+\varepsilon}\right)^{\lambda_k} \leq \beta_\varepsilon$

which is true since the LHS tends to zero as ${k \rightarrow \infty}$. Note that ${\beta_\varepsilon}$ depends only on ${\varepsilon}$ and ${\Lambda}$.

(b) Note that the uniform convergence of ${P_n}$ to ${f}$ implies that ${(P_n)}$ is Cauchy in the norm ${\|\cdot \|_\infty}$, and using (a) we have

$\displaystyle |a_{n,i}-a_{m,i}| \leq \beta_\varepsilon(1+\varepsilon)^{\lambda_i} \|P_n-P_m\|_\infty.$

As a consequence ${(a_{n,i})}$ is Cauchy in ${\Bbb{C}}$ for fixed ${i}$ and therefore it is convergent to some complex number ${a_i}$.

(c) If we denote ${g(x)=\sum_{n \geq 0}\alpha_nx^{\lambda_n}}$, then using the above relation for ${m \rightarrow \infty}$ we know that

$\displaystyle |P_n(x)-g(x)| \leq \sum_{i \geq 0} \beta_\varepsilon(1+\varepsilon)^{\lambda_i}x^{\lambda_i}\|P_n-f\|_\infty\leq \beta_\varepsilon \|P_n-f\|_\infty \frac{1}{1-(1+\varepsilon)x}.$

From the above inequality we can see that we have problems near ${x=1}$, but on the other hand, for every ${[0,b]\subset [0,1]}$ we can choose ${\varepsilon}$ small enough so that the last fraction is positive (and well defined on ${[0,b]}$) to see that ${|P_n(x)-g(x)|}$ is uniformly bounded by a constant times ${\|P_n-f\|_\infty}$ so ${P_n}$ converges uniformly to ${g}$ which was the desired result.

8. (a) If we assume that ${f_r}$ does not converge uniformly to ${f}$ then there exists ${(r_n) \rightarrow 1}$ such that ${\|f_{r_n}-f\|_\infty >c>0}$. As a consequence there exists a sequence ${(x_n)}$ of maximum values for ${|f_{r_n}-f|}$ on ${[0,1]}$ which (up to a subsequence) can be taken convergent to ${x_0}$. Therefore we have

$\displaystyle c< |f_{r_n}(x_n)-f(x_n)|=|f(r_nx_n)-f(x_n)|,$

but the last term of the inequality goes to ${0}$ as ${n\rightarrow \infty}$ by the continuity of ${f}$. This is a contradiction, and therefore the assumption made was false and

$\displaystyle \lim_{r \rightarrow 1}\|f_r-f\|_\infty=0.$

(b) Until here we have proven only that if ${f \in \overline M_\Lambda}$ then ${f}$ has the given form. We still have to prove the converse. Take

$\displaystyle f(x)=\sum_{i=0}^\infty a_i x^{\lambda_i}$

for ${x \in [0,1)}$. The key is (of course) to use point (a). Take the functions ${f_r(x)=f(rx)}$ which have the representation

$\displaystyle f_r(x)=\sum_{i=0}^\infty a_i r^{\lambda_i} x^{\lambda_i}.$

The functions ${f_r}$ are defined on ${[0,1]}$ but behave as ${f}$ on ${[0,r]}$ which is a compact subset of the domain of convergence. Therefore ${f_r}$ converges uniformly and so ${f_r \in \overline M_\Lambda}$. Apply (a) and we are done.