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.


\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 <e} 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<m})

6. (a) Use 3,4 and 5 (formula for the infimum, formula for the integral, split product in 3 parts {\lambda_n<m,\ m<\lambda_n \leq 2m,\ \lambda_n >2m}, 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.

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