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<b}) 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<t_{i+1}-t_i\leq \displaystyle\frac{1}{\mathcal{N}(S)}} (for {0\leq l <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.

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