Home > Analysis, Functional Analysis, Real Analysis > Agregation 2013 – Analysis – Part 4

## Agregation 2013 – Analysis – Part 4

Part IV: To be or not to be with separable dual

1. Consider ${S=[a,b]}$ a segment (${a) of ${[0,1]}$ and ${V}$ a closed linear subspace of ${C([0,1])}$ such that each function in ${V}$ is of class ${C^1}$ on ${S}$. For ${(x,y) \in S^2}$ such that ${x \neq y}$ we denote ${\displaystyle \xi_{x,y}(f)=\frac{f(x)-f(y)}{x-y}}$ where ${f \in V}$.

(a) Prove that ${\xi_{x,y} \in V^*}$.

(b) Prove that for every ${f \in V}$ we have

$\displaystyle \sup_{\substack{x,y \in S\\ x\neq y}} |\xi_{x,y}(f)|<\infty.$

(c) Deduce that there exists ${\mathcal{N}(S)>0}$ which verifies for each ${f \in V}$ and ${(x,y) \in S^2}$

$\displaystyle |f(x)-f(y)| \leq \mathcal{N}(S) |x-y|\|f\|_\infty.$

(d) Let ${(t_l)_{0\leq l \leq L}}$ be a finite sequence of points in ${S}$ such that ${0 (for ${0\leq l ) and ${t_0=a,t_L=b}$. Prove that

$\displaystyle \forall f \in V,\ \sup_{t \in S}|f(t)| \leq \sup_{0\leq l\leq L}|f(t_l)|+\frac{1}{2}\|f\|_\infty$

2. Let ${F_0}$ be a closed linear subspace of ${C([0,1])}$ such that every function in ${F_0}$ is of class ${C^1}$ on ${[0,1]}$. Prove that ${F_0}$ has finite dimension.

3. Consider ${X_0}$ a closed linear subspace of ${C([0,1])}$ such that every function of ${X_0}$ is of class ${C^1}$ on ${[0,1)}$.

(a) Give an example of such a subspace ${X_0}$ with infinite dimension. (hint: use part III).

Fix a sequence ${(a_j)}$ of ${[0,1)}$ strictly increasing towards ${1}$ ,a sequence ${(s_n)}$ of of ${[0,1]}$ strictly increasing towards ${1}$ and a sequence of integers ${n_j}$ strictly increasing towards ${\infty}$ such that ${a_0=s_0=0}$ and

$\displaystyle \forall j \in \Bbb{N} \text{ and } \forall n \in \{n_j,...,n_{j+1}-1\}, s_{n+1}-s_n \leq \frac{1}{\mathcal{N}([a_j,a_{j+1}])}$

(b) Let ${J:X_0 \rightarrow c}$ be an application which maps ${f}$ to ${(f(s_n))_{n \in \Bbb{N}}}$. Prove that this application is well defined and satisfies ${\|f\|_\infty \leq 2N_\infty(J(f)) \leq 2 \|f\|_\infty}$ for every ${f \in X_0}$.

(c) Deduce that ${X_0}$ is isomorphic to a subspace of ${c}$ and therefore ${X_0}$ is isomorphic to a subspace ${Z_0}$ of ${c_0}$.

(d) Justify that ${X_0^*}$ is isomorphic to ${Z_0^*}$ and therefore ${X_0^*}$ is separable. (hint: use Hahn-Banach)

4. We wish to prove by contradiction that ${C([0,1])^*}$ is not separable. Suppose that there exists a sequence ${(\omega_n)}$ which is dense in ${C([0,1])^*}$. For ${x \in [0,1]}$ and ${f \in C([0,1])}$ we denote ${\delta_x : f \mapsto f(x)}$ the evaluation on ${x}$.

(a) Justify that ${\delta_x \in C([0,1])^*}$ and that for every ${(x,y) \in [0,1]^2}$ with ${x \neq y}$ we have

$\displaystyle \|\delta_x -\delta_y\|_{C([0,1])^*}=2.$

Denote by ${\chi}$ the application which associates to ${x \in [0,1]}$ an integer ${n \in \Bbb{N}}$ such that ${\|\delta_x-\omega_n\|_{C([0,1])^*}<1}$.

(b) Prove that we define like this a one to one map from ${[0,1]}$ to ${\Bbb{N}}$ and conclude.

Hints: 1. (a) This is obvious, since ${\xi_{x,y}(f)}$ is just a linear combination of the evaluations of ${f}$ at ${x}$ and ${y}$.

(b) By the mean value theorem we know that there exists ${c}$ between ${x}$ and ${y}$ such that ${f'(c)=\xi_{x,y}(f)}$ and since ${f'}$ is continuous on ${S}$, for fixed ${f}$ ${f'(c)}$ is bounded above (for ${c \in S}$).

(c) Use Banach Steinhaus theorem for the family of linear maps ${\xi_{x,y}}$ defined on ${V}$ with values in ${\Bbb{R}}$ and deduce that

$\displaystyle \sup_{x,y \in S} \|\xi_{x,y}\|=\mathcal{N}(S) <\infty,$

which translates to

$\displaystyle |\xi_{x,y}(f)| \leq \mathcal{N}(S)\|f\|_\infty,$

exactly what we needed to prove.

(d) Pick ${f \in V}$. By (c) we can see that ${|f(t_{i+1})-f(t_i)|\leq \|f\|_\infty}$. Denote ${t^*}$ the point of maximum for ${f}$ on ${S}$ (which exists since ${S}$ is compact). Then ${t^*}$ is in an interval ${[t_i,t_{i+1}]}$ and it is closer to one side, WLOG suppose that it is closer to ${t_i}$. Then by (c) we have ${f(t^*)-f(t_i)\leq \frac{1}{2}\|f\|_\infty}$, which implies the desired result.

2. Choose ${N=\mathcal{N}([0,1])}$ and pick a sequence ${(t_i)}$ as in 1. (d). Next, suppose that ${f,g}$ are two functions in ${F_0}$ such that ${f(t_i)=g(t_i)}$. Note that ${f-g}$ is also in ${F_0}$ (because this is a vector subspace of ${C([0,1])}$, and the difference of two ${C^1}$ functions is ${C^1}$). Then the estimate in 1. (d) is valid for ${f-g}$. But that means

$\displaystyle \sup_{t \in [0,1]} |f(t)-g(t)|\leq \frac{1}{2}\|f-g\|_\infty,$

and therefore ${\|f-g\|_\infty=0}$, which means that ${f=g}$.

This means that each function ${f \in F_0}$ is uniquely determined by its values ${f(t_i), i=0..L}$. Therefore the application

$\displaystyle \mathcal{G} : F_0 \rightarrow \Bbb{R}^{L+1},\ \mathcal{G}(f)=(f(t_0),...,f(t_L))$

is an isomorphism and the conclusion follows.

3. (a) From part III one such space is ${\overline{M_\Lambda}}$ for ${\Lambda}$ such that ${\sum_{i \geq 0} 1/\lambda_i <\infty}$.

(b) Since ${(s_n) \rightarrow 1}$ we can see that ${f(s_n)}$ converges to ${f(1)}$ so the application is well defined. Furthermore, by 1. (d) we know that ${1/2 \|f\|_\infty \leq N_\infty(J(f))}$ from where we deduce that ${\|f\|_\infty \leq 2N_\infty(J(f))}$. The last inequality is trivial from the definitions of the sup norms.

(c) Using the same argument as in 2. we find that ${J}$ is injective, so ${X_0}$ is isomorphic to ${J(X_0)}$ which is a subspace of ${c}$. Use the application defined in Part I 3. to see that ${X_0}$ is isomorphic to ${Z_0=T(J(X_0))}$ which is a subspace of ${c_0}$.

(d) In general, if ${X,Y}$ are two isomorphic normed vector spaces (with an isomorphism ${T:X\rightarrow Y}$ with ${T^{-1}}$ continuous) then to each functional ${f: Y \rightarrow \Bbb{C}}$ we can associate the functional ${g \in X^*}$ by

$\displaystyle g(x)=f(Tx).$

This application defined on ${Y^*}$ with values in ${X^*}$ is linear, continuous and injective (by its definition and by the surjectivity of ${T}$).

Since ${T}$ is invertible, we can do the same thing the other way around: to ${g \in X^*}$ we associate ${f}$ by

$\displaystyle f(y)=g(T^{-1}y).$

We have thus constructed the inverse of the above linear continuous application, which means that ${X_0^*}$ is isomorphic to ${Z_0^*}$.

In order to deduce separability of ${X_0^*}$ we prove that ${c_0}$ is separable. To do that, a good candidate for a countable dense set is ${D=\{(q_n+ip_n) \subset (\Bbb{Q}+i\Bbb{Q})^\Bbb{N}, q_n=p_n =0,\ \forall n \geq n_0\}}$. Pick ${x_0 \in c_0}$ and ${\varepsilon>0}$. Then there exists ${n_0}$ such that for every ${n \geq n_0}$ we have ${|x_0^n|<\varepsilon}$. For all other ${n_0}$ first terms, we can find an ${\varepsilon}$-close number in ${D}$ so that there exists an element in ${D}$ at distance smaller than ${\varepsilon}$ of ${x_0}$. This means that ${\overline D=c_0}$.

Any subset of a separable space is also separable, so we are done. 4. (a) The evaluation ${\delta_x}$ is obviously a linear functional. If ${x \neq y}$ then we have

$\displaystyle \|\delta_x-\delta_y\|_{C([0,1])^*}=\sup_{\|f\|_\infty \leq 1} |\delta_x(f)-\delta_y(f)|=\sup_{\|f\|_\infty \leq 1} |f(x)-f(y)|\leq 2$

and we can construct a continuous function on ${[0,1]}$ such that ${\|f\|_\infty \leq 1 }$ and ${f(x)=1,\ f(y)=-1}$ (for example a piecewise linear interpolation). (b) If ${x\neq y}$ then if ${\|\delta_x-\omega_n\|<1}$ and ${\|\delta_y-\omega_m\|<1}$ we cannot have ${m=n}$. If ${m=n}$ then by (a) we have

$\displaystyle 2>\|\delta_x-\omega_n\|+\|\delta_y-\omega_n\|\geq \|\delta_x-\delta_y\|=2$

a contradiction. Therefore ${m \neq n}$ when ${x \neq y}$ and the application which associates to ${x}$ a positive integer ${n}$ is injective. This would mean that ${[0,1]}$ is countable, which is certainly not the case. As a consequence, our initial assumption was false and ${(C[0,1])^*}$ is not separable.