Math Challenge - July 2020

[mathwonk]

SPOILER-HINT for #5:

Take any two points a,b on an integral curve for the gradient flow of f. Then their distance apart along that curve seems to be f(b)-f(a), but is greater along any other path joining them. That seems to do it, at least based on an intuitive meaning of "geodesic".

 

[Infrared]

@mathwonk Yes, that's exactly the idea (and in fact shows that the integral curves are minimizing geodesics)

 

[mathwonk]

Spoiler/Hints for #9:

given any two points a,b in X and f:X-->X a surjective map with d(f(x),f(y)) ≤ d(x,y) for all x,y in X, choose a0 = f(a), b0 = f(b), a1 = a, b1 = b, a2: f(a2) = a1, b2: f(b2) = b1, a3: f(a3) = a2, b3: f(b3) = b2,...
an+1: f(an+1) = an, bn+1: f(bn+1) = bn,...

Show d(a0,b0) ≤ d(an,bn) all n. and that d(a0,an,) ≤ d(a1,an+1) all n.

Show that a0 is an accumulation point of the sequence an, n ≥ 1, and that the pair (a0,b0) is an accumulation point of the sequence (an,bn), n≥1. Conclude that d(a,b) ≤ d(f(a),f(b)).

 

[mathwonk]

Spoiler/hint: for #5:

Let c be a unit speed path, from time s to time t, joining two points of an integral curve for the gradient flow of f. Compare the integrals of <c',c'>, and <gradf, c'> from s to t, and think about what they tell you.

 

[mathwonk]

Hint for #1:In the weak topology, linear functions to R must be continuous, so if f is a real valued function, the inverse image of (-e,e) must be open. To be a topology, also finite intersections of such sets must be open. Also unions of such sets must be open, and that is all; hence that means the "smallest" open sets are finite intersections of such sets. Show that these sets all contain positive dimensional linear subspaces, hence are not norm-bounded if the space is infinite dimensional. E.g. a weak nbhd of 0 contains the intersection of the kernels of a finite sequence of linear functions, i.e. the kernel of a linear map to a finite dimensional space.

 

[julian]

I'm v.tired and about to hit the hay, so I'll just give this response to #5.

Write $X^a = g^{ab} \partial_b f$ and note that

\begin{align}
0 = \partial_c (1) = \nabla_c (g^{ab} \partial_a f \partial_b f) = 2 X^b \nabla_c \partial_b f = 2 X^b \nabla_b \partial_c f = 2 X^b \nabla_b X_c
\nonumber
\end{align}

where we have used $\nabla_c \partial_b f = \partial_c \partial_b f - \Gamma^d_{cb} \partial_d f = \partial_b \partial_c f - \Gamma^d_{bc} \partial_d f = \nabla_b \partial_c f$ and $\nabla_c g^{ab} = 0$ (I'm using $\nabla_a$ to mean the covariant derivative). We have obtained

\begin{align}
X^b \nabla_b X^a = 0 .
\nonumber \\
\end{align}

(where we have used $\nabla_c g^{ab} = 0$ again). An integral curve $x^a = x^a(u)$ of a vector field is a curve such that

\begin{align}
\frac{d x^a (u)}{du} = X^a (x^b (u)) .
\nonumber \\
\end{align}

Substituting this into $X^b \nabla_b X^a = 0$ gives

\begin{align}
0 & = X^b \nabla_b X^a
\nonumber \\
& = \frac{d x^b (u)}{du} \nabla_b \left( \frac{d x^a (u)}{du} \right)
\nonumber \\
& = \frac{d x^b (u)}{du} \left\{ \frac{\partial}{\partial x^b} \left( \frac{d x^a (u)}{du} \right) + \Gamma^a_{bc} \frac{d x^c (u)}{du} \right\}
\nonumber \\
& = \frac{d^2 x^a (u)}{du^2} + \Gamma^a_{bc} \frac{d x^b (u)}{du} \frac{d x^c (u)}{du}
\nonumber
\end{align}

which is the metric geodesic equation obtained by minimizing the distance between two fixed points (with $u$ an affine parameter).

 

[member 587159]

Show that these sets all contain positive dimensional linear subspaces, hence are not norm-bounded if the space is infinite dimensional.

How would you show this?

 

[wrobel]

Regarding #5 another way to get the result is to apply the vector field straightening theorem to the vector field

$$g^{ij}\frac{\partial f}{\partial x^j}$$

 

[wrobel]

Spoiler/Hints for #9:

given any two points a,b in X and f:X-->X a surjective map with d(f(x),f(y)) ≤ d(x,y) for all x,y in X, choose a0 = f(a), b0 = f(b), a1 = a, b1 = b, a2: f(a2) = a1, b2: f(b2) = b1, a3: f(a3) = a2, b3: f(b3) = b2,...
an+1: f(an+1) = an, bn+1: f(bn+1) = bn,...

Show d(a0,b0) ≤ d(an,bn) all n. and that d(a0,an,) ≤ d(a1,an+1) all n.

Show that a0 is an accumulation point of the sequence an, n ≥ 1, and that the pair (a0,b0) is an accumulation point of the sequence (an,bn), n≥1. Conclude that d(a,b) ≤ d(f(a),f(b)).

seems to be ok

 

[nuuskur]

Spoiler: Problem 1

Assume for a contradiction $\sigma (V,V')$ is metrisable. The Banach space $V'$ is infinite dimensional. Since $\sigma (V,V')$ is a metrisable locally convex topology, its neighborhood basis of zero is generated by a sequence of seminorms $p_{\varphi _n}$, where $\varphi _n\in V', n\in\mathbb N$. Wo.l.o.g this basis can be expressed as
$$S_{n} := \bigcap _{k=1}^n \varphi _k^{-1} (B(0,1)),\quad n\in\mathbb N.$$
Take $\varphi \in V'$. Fix $\varepsilon >0$. Due to continuity w.r.t $\sigma (V,V')$ we have $S_{N_\varepsilon} \subseteq \varphi ^{-1}(B(0,\varepsilon))$ for some $N_\varepsilon\in\mathbb N$. Put $N := \min _\varepsilon N_\varepsilon$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _k$ implies $x\in \mathrm{Ker}\varphi$. Equivalently, $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$. Thus, $V'$ admits a countable spanning set, which is impossible. Therefore, $\sigma (V,V')$ cannot be metrisable.

 

[member 587159]

Spoiler: Problem 1

Assume for a contradiction $\sigma (V,V')$ is metrisable. The Banach space $V'$ is infinite dimensional. Since $\sigma (V,V')$ is a metrisable locally convex topology, its neighborhood basis of zero is generated by a sequence of seminorms $p_{\varphi _n}$, where $\varphi _n\in V', n\in\mathbb N$. Wo.l.o.g this basis can be expressed as
$$S_{n} := \bigcap _{k=1}^n \varphi _k^{-1} (B(0,1)),\quad n\in\mathbb N.$$
Take $\varphi \in V'$. Fix $\varepsilon >0$. Due to continuity w.r.t $\sigma (V,V')$ we have $S_{N_\varepsilon} \subseteq \varphi ^{-1}(B(0,\varepsilon))$ for some $N_\varepsilon\in\mathbb N$. Put $N := \min _\varepsilon N_\varepsilon$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _k$ implies $x\in \mathrm{Ker}\varphi$. Equivalently, $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$. Thus, $V'$ admits a countable spanning set, which is impossible. Therefore, $\sigma (V,V')$ cannot be metrisable.

I'm aware that that every locally convex metrizable topological vector space has topology generated by a countable family of seminorms. However, you seem to be assuming that these seminorms are induced by functionals (given a functional $\omega: V \to \Bbb{K}$, its modulus $|\omega|: V \to [0,\infty[$ gives a seminorm). Can you explain why this is true or give a reference to that result?

 

[nuuskur]

I'm aware that that every locally convex metrizable topological vector space has topology generated by a countable family of seminorms. However, you seem to be assuming that these seminorms are induced by functionals (given a functional $\omega: V \to \Bbb{K}$, its modulus $|\omega|: V \to [0,\infty[$ gives a seminorm). Can you explain why this is true or give a reference to that result?

Given the dual pair $(V,V')$, its weak topology $\sigma (V,V')$ is generated by the family of seminorms $p_f : x\mapsto |f(x)|,\ f\in V'$. If the topology is metrisable, this family may be assumed to be at most countable.

 

[Infrared]

@julian This looks right, but I think the coordinate-free approach suggested by @mathwonk is a bit cleaner (and also has the advantage of showing that the integral curves are minimizing geodesics).

@wrobel Could you sketch your argument with the vector field straightening theorem?

 

[mathwonk]

Spoiler #5:Everything follows just from looking at the integral of <gradf, c'>, where c is a unit speed curve.

summary; | integral of <gradf,c'> dt| ≤ |t-s| = length of path c, with equality iff gradf = ± c'.

 

[wrobel]

@l Could you sketch your argument with the vector field straightening theorem?

There are local coordinates such that $g^{ij}\frac{\partial f}{\partial x^j}=\delta_1^i$ thus
$$\frac{\partial f}{\partial x^j}=g_{1j};$$ and
$$|\nabla f|=1\Longrightarrow g_{11}=1\Longrightarrow\frac{\partial f}{\partial x^1}=1\Longrightarrow \Gamma_{11}^i=0.$$
Furthermore,
$$\dot x^i=\delta^i_1\Longrightarrow x^i(t)=\delta^i_1t+x^i_0.$$ This function obviously satisfies the equation
$$\ddot x^i+\Gamma_{ks}^i\dot x^k\dot x^i=0$$

 

[mathwonk]

@Math_QED: post #110 edited to include more argument.

 

[member 587159]

Spoiler: Problem 1

Assume for a contradiction $\sigma (V,V')$ is metrisable. The Banach space $V'$ is infinite dimensional. Since $\sigma (V,V')$ is a metrisable locally convex topology, its neighborhood basis of zero is generated by a sequence of seminorms $p_{\varphi _n}$, where $\varphi _n\in V', n\in\mathbb N$. Wo.l.o.g this basis can be expressed as
$$S_{n} := \bigcap _{k=1}^n \varphi _k^{-1} (B(0,1)),\quad n\in\mathbb N.$$
Take $\varphi \in V'$. Fix $\varepsilon >0$. Due to continuity w.r.t $\sigma (V,V')$ we have $S_{N_\varepsilon} \subseteq \varphi ^{-1}(B(0,\varepsilon))$ for some $N_\varepsilon\in\mathbb N$. Put $N := \min _\varepsilon N_\varepsilon$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _k$ implies $x\in \mathrm{Ker}\varphi$. Equivalently, $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$. Thus, $V'$ admits a countable spanning set, which is impossible. Therefore, $\sigma (V,V')$ cannot be metrisable.

I guess the idea is correct, but it is unclear to me why you work with the $\epsilon$ and the minimum? Doesn't the following work instead?

Let $\varphi \in V^*$. Then there exists $n$ with $S_n = \bigcap_{k=1}^n \varphi_k^{-1}(B(0,1)) \subseteq \varphi^{-1}(B(0,1))$. This implies that $\bigcap_{k=1}^n \ker \varphi_k \subseteq \ker \varphi$: indeed if $\varphi_k(x)=0$ for $k=1, \dots, n$, then also $\varphi_k(tx)=0$ for all $t\in \Bbb{R}$ and thus $tx\in S_n\subseteq \varphi^{-1}(B(0,1))$, so that $t\varphi(x) \in B(0,1)$. Since this holds for all $t \in \Bbb{R}$, this forces $\varphi(x)=0$. You then can conclude like you did.

I think it is non-trivial that you can choose the basis like you did: I agree that the topology $\sigma(V,V')$ is generated by the seminorms $\{p_f: f \in V^*\}$. But then the following two questions remain:

(1) How does metrizability imply that you can choose $(f_n)_n$ such that the family of seminorms $\{p_{f_n}: n \geq 1\}$ still generate the topology? (Provide proof or a reference).

(2) Why can we choose the basis in this particular form: i.e. as inverse images of the unit ball?

 

[benorin]

Summary:: Functional Analysis, Topology, Differential Geometry, Analysis, Physics
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

9. Let $(X,d)$ be a compact metric space and $f:X\to X$ be a mapping onto. Assume that $d(f(x),f(y))\le d(x,y),\quad \forall x,y\in X.$ Show that $d(f(x),f(y))= d(x,y),\quad \forall x,y\in X.$ (WR)

This is what I got so far, but my expertise here is lacking to say the least, if you've any skills here please correct what I've got (or anybody really) as Topology is really new to me, thanks.

Spoiler: What I got so far

Work: $f:X\to X$ is continuous by Munkres' Topology Theorem 21.1, pg 129 which is the $\epsilon -\delta$ definition of continuity of $f:X\to Y$ for metric spaces $X$ and $Y$ with respective metrics $d_X$ and $d_Y$, here taking $X=Y$ and metrics $d_X (x,y) = d_Y (x,y) :=d(x,y)$ since given $x\in X$ and $\epsilon >0, \forall n\in \mathbb{N}$ choose $\delta_n = \epsilon +\tfrac{1}{n}$ such that $d_X(x,y)<\delta _n = \epsilon +\tfrac{1}{n}\implies d_Y(f(x), f(y)) \leq d_X (x,y) < \epsilon +\tfrac{1}{n}\implies d_X(f(x),f(y)) < \epsilon$ where the business with adding the $\tfrac{1}{n}$ term I'm not certain is quite right but was intended to deal with the strictness of the inequalities.

 

[wrobel]

ork: is continuous by Munkres' Topology Theorem 21.1, pg 129 which

sure it is continuous ,moreover it is Lipschitz

 

[nuuskur]

I guess the idea is correct, but it is unclear to me why you work with the $\epsilon$ and the minimum? Doesn't the following work instead?

Spoiler

Let $\varphi \in V^*$. Then there exists $n$ with $S_n = \bigcap_{k=1}^n \varphi_k^{-1}(B(0,1)) \subseteq \varphi^{-1}(B(0,1))$. This implies that $\bigcap_{k=1}^n \ker \varphi_k \subseteq \ker \varphi$: indeed if $\varphi_k(x)=0$ for $k=1, \dots, n$, then also $\varphi_k(tx)=0$ for all $t\in \Bbb{R}$ and thus $tx\in S_n\subseteq \varphi^{-1}(B(0,1))$, so that $t\varphi(x) \in B(0,1)$. Since this holds for all $t \in \Bbb{R}$, this forces $\varphi(x)=0$. You then can conclude like you did.

It does work. I don't really understand your objection, though. Similarly, one could ask why you would work with the following if some other method is also sufficient. I think it's apples and oranges. I might be wrong.

I think it is non-trivial that you can choose the basis like you did: I agree that the topology $\sigma(V,V')$ is generated by the seminorms $\{p_f: f \in V^*\}$. But then the following two questions remain:

(1) How does metrizability imply that you can choose $(f_n)_n$ such that the family of seminorms $\{p_{f_n}: n \geq 1\}$ still generate the topology? (Provide proof or a reference).

(2) Why can we choose the basis in this particular form: i.e. as inverse images of the unit ball?

Spoiler: (2)

Nh basis of zero for $\sigma (V,V')$ is given by the family
$$B_{\varepsilon, f_1,\ldots,f_n} := \left\{ x\in V \mid \max\limits_{1\leq j\leq n} |f_j(x)| \leq \varepsilon \right\},\quad \varepsilon > 0,\quad f_1,\ldots,f_n\in V',\quad n\in\mathbb N.$$
Since $V'$ is a vector space, we can instead take $\frac{1}{\varepsilon}f_j$ and obtain a basis
$$B_{1,f_1,\ldots,f_n} = \left\{ x\in V \mid \max\limits _{1\leq j\leq n} |f_j(x)| \leq 1 \right\} = \bigcap _{k=1}^n f_k^{-1}(B(0,1)),\quad f_1,\ldots,f_n\in V',\quad n\in\mathbb N.$$
The second equality is clear, I think.

Spoiler: (1)

Given a locally convex topology its nh basis of zero can be assumed to consist of closed absolutely convex subsets. Let the topology be metrisable, then this basis can be assumed to be countable. Suppose such a basis is $S_n,\ n\in\mathbb N$. Its generating family of seminorms can be picked as the Minkowski functionals $p_{S_n}$ of basis elements. By Hahn-Banach, there exists $\varphi _n\in V'$ s.t $|\varphi _n | \leq p_{S_n},\quad n\in\mathbb N$.

More generally, nh basis of zero generated by seminorms is given as
$$B_{\varepsilon, n} = \left\{ x\in V \mid \max\limits_{1\leq j\leq n} p_{S_j}(x) \leq \varepsilon \right\},\quad \varepsilon >0,\quad n\in\mathbb N.$$
Now, given $n\in\mathbb N$ and $\varepsilon >0$, we can find $\varepsilon _0 >0$ s.t
$$\left\{ x\in V \mid \max\limits_{1\leq j\leq n} |\varphi _n(x)| \leq \varepsilon _0 \right\} \subseteq B_{\varepsilon,n}.$$
Thus, the sequence of functionals $\varphi _n$ also generates a nh basis of zero.

I appreciate the scrutiny. Don't let me get away with handwaving.

 

[etotheipi]

Spoiler: Problem 12

12. Given a positive integer in decimal representation without zeros. We build a new integer by concatenation of the number of even digits, the number of odd digits, and the number of all digits (the sum of the former two). Then we proceed with that number.
Determine whether this algorithm always comes to a halt. What is or should be the criterion to stop?

I don't really know if this is what you're after, but for any given integer with 4 or more digits there will be either at least 2 even digits or at least 2 odd digits, and the next pass of the algorithm will always reduce the length of that string by at least one digit. You can just keep doing this until you get down to 3 digits, which are either

$\{\text{even}, \text{even}, \text{even}\} \implies 303 \implies 123$
$\{\text{even}, \text{even}, \text{odd}\} \implies 213 \implies 123$
$\{\text{even}, \text{odd}, \text{odd}\} \implies 123$
$\{\text{odd}, \text{odd}, \text{odd}\} \implies 033 \implies 123$

And I guess if you start off with an integer with 1 or 2 digits, the next pass gives you a 3 digit number and we're back to the above.

 

[fresh_42]

Spoiler: Problem 12

I don't really know if this is what you're after, but for any given integer with 4 or more digits there will be either at least 2 even digits or at least 2 odd digits, and the next pass of the algorithm will always reduce the length of that string by at least one digit. You can just keep doing this until you get down to 3 digits, which are either

$\{\text{even}, \text{even}, \text{even}\} \implies 303 \implies 123$
$\{\text{even}, \text{even}, \text{odd}\} \implies 213 \implies 123$
$\{\text{even}, \text{odd}, \text{odd}\} \implies 123$
$\{\text{odd}, \text{odd}, \text{odd}\} \implies 013 \implies 123$

And I guess if you start off with an integer with 1 or 2 digits, the next pass gives you a 3 digit number and we're back to the above.

I had a more formal argument in mind, but ok. I think you mistyped 013. Also I guess is a bit thin. And you could have actually answered the questions. However, these are only formal deficits.

 

[nuuskur]

Point of http://ssdnm.mimuw.edu.pl/pliki/wyklady/PlebanekAbstrakt_us.pdf

Spoiler: Problem 4

Assume for a contradiction $C[0,1]$ is isomorphic to a dual Banach space.

Take a sequence of functions with norm $1$ in $C[0,1]$ and with pairwise disjoint support. For instance, pick a sequence of pairwise disjoint intervals in $[0,1]$ and pick a function whose support is that interval. The linear span of this sequence is dense in $c_0$: match the sequence with the canonical basis in $c_0$.

The space $C[0,1]$ is separable. By Theorem 3.1 (Sobczyk), the closed copy of $c_0$ is complemented in $C[0,1]$. By assumption $c_0$ is complemented in a dual Banach space. Dual Banach spaces are complemented in their second dual, therefore $c_0$ is complemented in its second dual, which is $\ell _\infty$. But this contradicts Theorem 3.2 (Phillips-Sobczyk), by which $c_0$ can't be complemented in $\ell _\infty$. Thus, $C[0,1]$ can't be isomorphic to a dual Banach space.

 

[wrobel]

I have already stuck at here:

Take a sequence of functions with norm in and with pairwise disjoint support. For instance, pick a sequence of pairwise disjoint intervals in and pick a function whose support is that interval. The linear span of this sequence is dense in : match the sequence with the canonical basis in .

please provide details

 

[nuuskur]

I have already stuck at here:
please provide details

Spoiler

Firstly, we want the sequence to be linearly independent, hence pairwise disjoint supports. So, denote it $f_n\in C[0,1],\ n\in\mathbb N$. We also want an isometry, so pick the $f_n$ with norm $1$ i.e make sure $f_n$ reaches the unit sphere. Let the $f_n$ span the subspace $K\subseteq C[0,1]$. Then put $\varphi (f_n) = e_n\in c_0$ where $e_n = (\underbrace{0,\ldots,1}_{n},0,\ldots),\ n\in\mathbb N$. The disjointness of supports guarantees we have an isometry.
Let $a:=\sum _{k=1}^\infty \lambda _ke_k\in c_0$ and $\varepsilon >0$. Take $N\in\mathbb N$ such that $\left \|\sum _{k>N} \lambda _ke_k \right \| <\varepsilon$. So $a$ is close to $\varphi \left ( \sum _{k=1}^N \lambda _kf_k \right )$. Now $\overline{\varphi (K)} = c_0$ implies $\overline{K} \cong c_0$.

 

[wrobel]

ok
please explain the next step:

By assumption is complemented in a dual Banach space.

 

[nuuskur]

ok
please explain the next step:

Spoiler

Remark. Formally, the terminology is ".. is a complemented subspace of ..", but I'm used to saying " .. is complemented in ..".

Let a closed subspace $K\subseteq X^*$ be complemented in $X^*$. In other words, there exist continuous maps $\alpha : K \to X^*$ and $\beta : X^* \to K$ such that $\beta\circ \alpha = \mathrm{id}_{K}$. Denote $j_X :X\to X^{**}$ the canonical embedding, then we have a projection $j_{K}\circ\beta\circ (j_X)^*\circ\alpha ^{**}$ onto $j_K(K)\cong K$. One can show the following diagram (there's a typo - it should read $\beta\circ \alpha = \mathrm{id}_K$) is commutative

Spoiler

..it's a routine check, but it takes some time and I'm too tired right now.

 

[nuuskur]

Spoiler

Remark. Formally, the terminology is ".. is a complemented subspace of ..", but I'm used to saying " .. is complemented in ..".

Let a closed subspace $K\subseteq X^*$ be complemented in $X^*$. In other words, there exist continuous maps $\alpha : K \to X^*$ and $\beta : X^* \to K$ such that $\beta\circ \alpha = \mathrm{id}_{K}$. Denote $j_X :X\to X^{**}$ the canonical embedding, then we have a projection $j_{K}\circ\beta\circ (j_X)^*\circ\alpha ^{**}$ onto $j_K(K)\cong K$.

Alright, I'm recharged now.

Spoiler

Spoiler: diagram

This diagram is commutative. In particular $(j_X)^* \circ \alpha ^{**} \circ j_K = \alpha$, therefore $j_K \circ \beta \circ (j_X)^* \circ \alpha ^{**}$ is a projection onto $j_K(K) \cong K$ i.e $K$ is complemented in its second dual. It suffices to check $\alpha^{**}\circ j_K = j_{X^*} \circ \alpha$, because the transpose $(j_X)^*$ is left inverse to the embedding $j_{X^*}$ (the bottom subtriangle commutes). Fix $k\in K$ and $x^{**}\in X^{**}$. Then we have the equalities
$$\begin{align*} (j_{X^*} \circ \alpha)(k)(x^{**}) &=j_{X^*}(\alpha (k))(x^{**}) \\ &= x^{**}(\alpha(k)) \\ &=\alpha ^* (x^{**})(k) \\ &= j_K(k)(\alpha ^*(x^{**})) \\ &=\alpha ^{**}(j_K(k))(x^{**}) \\ &= (\alpha ^{**} \circ j_K)(k)(x^{**}). \end{align*}$$

In particular, every dual Banach space is complemented in its second dual. This fact is also directly proved in the notes I linked in #128.

 

[wrobel]

By assumption is complemented in a dual Banach space.

In which dual space c_0 is complemented? By which assumption?

 

[nuuskur]

In which dual space c_0 is complemented? By which assumption?

The assumption is $C[0,1]$ (which is separable) is isomorphic to a dual Banach space $X^*$. By Sobczyk's theorem $c_0$ is complemented in $X^* \cong C[0,1]$.

 

[wrobel]

Oh now I see. It looks correct. But by the Krein–Milman theorem it would be easier

 

[nuuskur]

I understand the idea, but I haven't worked through a proof of Krein-Milman, so it's a bit uncomfortable language. On the other hand, I am very comfortable with diagram chasing :)

@Math_QED Perhaps, you've been busy or you just missed #125.

 

[member 587159]

@Math_QED Perhaps, you've been busy or you just missed #125.

Yes, sorry for the delay in a couple of days I will have time to look at your post.