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2 Differentiable open maps

Definition 2.1. A map F : X Y between two Banach spaces is said to be (Fr´echet) → differentiable at a point x X if there exists a (unique) bounded linear operator JF (x) from ∈ X into Y , called the derivative of F at x, such that

F (x + h) F (x)= JF (x)h + o( h ) (h 0). − || || → Furthermore, x is said to be a regular (critical) point of F if JF (x) has (has not) an inverse in B(Y, X), the space of bounded linear operators from Y to X; and F is said to be C1 if it is differentiable everywhere and x JF (x) is a continuous map from X to B(X,Y ). A map 7→ between two topological spaces is said to be open if it maps open sets to open sets. We now make an initial study of maps of the form (1.1). Take a positive constant C such that x C x for all x X. For any h X 0 , ||| ||| ≤ k k ∈ ∈ \{ } F (h) F (0) F (h) 1 1 k − k = k k = exp( ) exp( ), h h − h s ≤ −Cs h s k k k k ||| ||| k k where the last term goes to zero as h 0 in (X, )). According to Definition 2.1, F is → k · k differentiable at the origin with derivative JF (0) = 0, the zero element of B(X) := B(X, X). Thus x = 0 is a critical point of F . Theorem 2.2. The map F defined by (1.1) is not open. Proof. If this was not true, then by considering F (0) = 0, there exists a δ (0, 1) such that ∈ for any given y X with y = 1, one can find an element xy in the unit open ball in (X, ) ∈ k k k·k such that δ 1 y = F (xy) = exp( s ) xy. 2 − xy · ||| ||| Obviously, xy must be of the form xy = ryy for some ry (0, 1). Consequently, ∈ δ 1 1 = exp( ) ry exp( ), 2 −rs y s · ≤ − y s y||| ||| ||| ||| which implies that 1 1 s y . ||| ||| ≥ ln 2  δ  Therefore, is stronger than on X. It is assumed that is strictly weaker than ||| · ||| k · k ||| · ||| , so a contradiction is derived. This finishes the proof of Theorem 2.2. k · k Obviously, the map F defined by (1.1) is a bijective map on X. The proof of Theorem 2.2 implies that the inverse F −1 of F is not continuous at F (0) = 0 of X. We now can make a slightly stronger estimate as follows. Since is strictly weaker than , one can find ||| · ||| 1k · k yn elements yn X for large enough n N, such that yn = 1 and yn = . Define zn = , ∈ ∈ k k ||| ||| n n and let xn X be such that F (xn) = zn. Obviously, xn must be of the form xn = γnyn for ∈ some γn > 0. Hence ns 1 exp s γn = F (xn) = zn = , − γn · k k k k n   ns or equivalently nγn = exp s , from which one can easily deduce that γn √n for sufficiently γn ≥ −1 large n. To conclude, we see that zn 0 while F (zn) as n goes to infinity. k k→ k k→∞ Differentiable maps with isolated critical points 3

3 First example

Consider ∞ N l2 = x = (x ,x , ) R : x2 < 1 2 · · · ∈ k ∞ n Xk=1 o 1 1 ∞ 2 ∞ 2 2 2 xk 2 with standard norm x = xk . Take x = k , another norm on l , which || || k=1 ||| ||| k=1 is strictly weaker than .  P   P  k · k 1 1 2 Theorem 3.1. F (x) = exp( 2 ) x is a C map on (l , ) with a unique critical point − |||x||| · k · k at the origin.

Proof. In Section 2 we have shown that F is differentiable at the origin with derivative JF (0) = 0. In the following we will first show that F is also differentiable on l2 0 . Let x l2 0 \{ } ∈ \{ } and h l2 be fixed such that h < |||x|||. For any N N, denote ∈ || || 2 ∈ N N x = (x1,x2,...,xN , 0, 0,...), h = (h1, h2,...,hN , 0, 0,...).

N We always let N be large enough so that hN < |||x ||| and |||x||| < xN . For such N, || || 2 2 ||| ||| 1 1 fN : t [0, 1] − = − R xN thN 2 N 2 ∈ 7→ + (xk+thk) ∈ ||| ||| k k=1 P is a smooth function, and 1 16 fN L∞ := sup fN (t) = sup || || | | xN + thN 2 ≤ x 2 t∈[0,1] t∈[0,1] ||| ||| ||| ||| since for any t [0, 1], ∈ xN x xN + thN xN hN xN hN ||| ||| || ||. ||| ||| ≥ ||| ||| − ||| ||| ≥ ||| ||| − || || ≥ 2 ≥ 4 Moreover, direct computation shows that

N 2(x + th ) f ′ (t)= f 2 (t) k k h , N N k k Xk=1 N N 2 ′′ 2(x + th ) 2 2h f (t) = 2f 3 (t) k k h + f 2 (t) k . N N k k N k  Xk=1  Xk=1 Hence 2 ′ 16 f L∞ 3 x h , k N k ≤ x 4 · k k · k k ||| ||| 3 2 ′′ 16 2 16 2 f L∞ (2 9 x + 2) h . k N k ≤ · x 6 · k k x 4 · · k k ||| ||| ||| ||| 4 C. Feng and L. Li

More briefly, there exists a positive constant M depending only on x such that

i d i fN M h (3.1) dti L∞ ≤ k k

for i = 0, 1, 2. As a consequence, di efN M h i (3.2) dti L∞ ≤ k k

for i = 0, 1, 2, where M is another positive constant c depending only M. Note that 1 1 F (x + h) F (x) = expc (x + h) exp x − − x + h 2 · − − x 2 ·  ||| |||   ||| |||  = lim efN (1) efN (0) x + efN (1) h N→∞ − · ·  d 1 d2 d efN (t) efN (t) x efN (0) efN (t) h, = lim + 2 + + N→∞ dt t=0 2 dt t=αN · dt t=βN ·     where the existence of αN (0, 1) and βN (0, 1) is ensured by Taylor’s theorem in calculus. ∈ ∈ Considering (3.2), one gets

d M x F (x + h) F (x) lim efN (t) x lim efN (0) h k k + M h 2, − − N→∞ dt t=0 · − N→∞ · ≤ 2 k k   c where c 1 lim efN (0) = exp( ), N→∞ − x 2 ||| ||| ∞ d 1 1 2x h lim efN (t) = exp( ) k k . N→∞ dt t=0 − x 2 · x 4 · k ||| ||| ||| ||| k=1 X According to the Cauchy-Schwarz inequality, the linear map

∞ 1 1 2x h h l2 exp( ) k k x l2 ∈ 7→ − x 2 · x 4 · k · ∈ ||| ||| ||| ||| Xk=1 is a bounded operator on (l2, ), where now h can represent an arbitrary element of l2. By k · k Definition 2.1, F is differentiable at x with derivative given by

∞ 1 1 2xkhk 1 JF (x)h = exp( ) x + exp( ) h. (3.3) − x 2 · x 4 · k · − x 2 · ||| ||| ||| ||| Xk=1 ||| ||| 2 2 It is easy to verify by composition rule that x JF (x) is a continuous map from l to B(l ), 7→ so F is C1. 2 Assuming x is an arbitrary non-zero element of l , we will prove next that JF (x) is bijective on l2. For any y l2, we want to find an h l2 such that ∈ ∈ ∞ 1 1 2x h 1 exp( ) k k x + exp( ) h = y, − x 2 · x 4 · k · − x 2 · ||| ||| ||| ||| Xk=1 ||| ||| Differentiable maps with isolated critical points 5 or equivalently ∞ 1 2x h 1 k k x + h = exp( ) y =:y. ˜ x 4 · k · x 2 · ||| ||| Xk=1 ||| ||| Thus h must be of the form h =y ˜ + γx for some γ R, from which we deduce ∈ ∞ 1 2x (˜y + γx ) k k k x + γx = 0. x 4 · k · ||| ||| Xk=1 As x = 0, we get 6 ∞ 1 2xky˜k 4 |||x||| · k γ = k=1 . P2 − 1+ |||x|||2 Hence there exists a unique solution

∞ 1 2xkyk |||x|||4 k 1 · k=1 h = exp( 2 ) y 2 x . (3.4) x · − 1+ P 2 · ||| ||| " |||x||| #

In particular, if y = 0 then the solution must be h = 0. This proves that JF (x) is both surjective and injective. Finally, it follows from Banach’s isomorphism theorem that JF (x) has an inverse in B(l2). In other words, x is a regular point of F . This finishes the proof.

According to Theorem 2.2, the map given by Theorem 3.1 is our first counterexample to Saint Raymond’s question.

Remark 3.2. We remark that (3.4) can also be written as

1 4 J 2(x)y 1 |||x||| · |||·||| h = exp( 2 ) y 1 x . x · − 1+ 4 J 2 (x)x · ||| ||| " |||x||| · |||·||| # Remark 3.3. Following the notations in Theorem 3.1, we remark that 1 x l2 exp( ) x l2 ∈ 7→ − x 4 · ∈ ||| ||| is another counterexample to Saint Raymond’s question. The continuous differentiability of 1 this map is much easier to be deduced by composition rule since as t exp( 2 ) is a smooth 7→ − t function on R, it thus remains only to show that x x 2 is a C1 function on l2. Similarly, 1 7→ ||| ||| 2 1 considering t exp( t ) is smooth on [0, ) and after showing is a non-negative C 27→ − ∞ 1 ||| · |||1 2 function on l , we immediately get that x exp( 2 ) is also a C function on l . 7→ − |||x||| 6 C. Feng and L. Li

4 Second example

Consider ∞ p N p l = x = (x ,x , ) R : xk < (1 p< ) 1 2 · · · ∈ | | ∞ ≤ ∞ k n X=1 o ∞ 1 p p with standard norm x p = xk . Given arbitrary 1 p1

x p2 x p1 (4.1) k k ≤ k k p1 p1 for all x l . Hence p2 , another norm on l , is strictly weaker than p1 . ∈ k · k k · k q 1 Theorem 4.1. Assume q p and q is an even integer. Then G(x) := x q is a C function p ≥ || || on (l , p). k · k Proof. Note q is an even integer, so for any x, h lp, ∈ G(x + h) G(x)= x + h q x q − || ||q − || ||q ∞ q q = (xk + hk) x − k k=1 h i X q ∞ q = hj xq−j j k k k j X=1 X=1   q ∞ ∞ q = q h xq−1 + hj xq−j k k j k k j Xk=1 X=2 Xk=1   =: I + II.

1 1 ′ q We will estimate I and II in the following. For I, by H¨older’s inequality ( q + q′ = 1, q = q−1 is the conjugate index of q) and (4.1) (p = p q = p ), we have 1 ≤ 2 ∞ q−1 q−1 q−1 hkxk h q x q h p x p , ≤ || || · || || ≤ || || · || || k=1 X so ∞ p q−1 h l q hkx R ∈ 7→ k ∈ k X=1 p is a bounded linear functional on (l , p). Similarly, for II, k · k q ∞ q q q j q−j q j q−j q j q−j hkxk h q x q h p x p . j ≤ j || || || || ≤ j || || || || j=2 k=1   j=2   j=2   X X X X Therefore, G is differentiable at x lp with derivative given by ∈ ∞ q−1 2 JG(x)h = q hkx (h l ). k ∈ Xk=1 Differentiable maps with isolated critical points 7

p Next, we will show that x JG(x) is a continuous map from (l , p) to its dual space. For 7→ k · k any x,z,h lp, ∈ ∞ q−1 q−1 JG(x + z)h JG(x)h = q hk (xk + zk) x − − k k X=1 h i q−1 ∞ q 1 = q h − zi xq−1−i k i k k i k X=1 X=1   ∞ q−2 ∞ q 1 = q h zq−1 + q − h zi xq−1−i k k i k k k k i k X=1 X=1   X=1 =: III + IV.

Similar to the method of estimating I and II, we have for III and IV (applying generalized 1 i q−1−i H¨older’s inequality to IV with q + q + q = 1 for each individual term),

∞ q−1 q−1 q−1 hkzk h q z q h p z p , ≤ || || · || || ≤ || || · || || Xk=1 ∞ i q−1−i i q−1−i i q−1−i hkzkxk h q z q x q h p z p x p (1 i q 2). ≤ || || · || || · || || ≤ || || · || || · || || ≤ ≤ − k=1 X

Consequently, q−2 q−1 q 1 i q−1−i JG(x + z) JG(x) q z + q − z x . || − || ≤ || ||p i || ||p|| ||p i X=1   p Hence, x JG(x) is a continuous map from (l , p) to its dual space. To conclude, G is a 1 7→ p k · k C function on (l , p). This finishes the proof. k · k Similar to the discussions in Sections 2 and 3 as well as considering Remark 3.3 and Theorem 4.1, it is not difficult to show that 1 F : x exp( q ) x 7→ − x q · k k p 1 on (l , p) is a non-open C map with a unique critical point at the origin as long as q>p k · k p (so that q is strictly weaker than p on l ) is an even integer. In particular, the solution p k · k p k · k h l to JF (x)h = y, y l , is uniquely given by (see also Remark 3.2) ∈ ∈ ∞ q q−1 2q x yk kxk k 1 q · k=1 h = exp( q ) y q x x q · − 1+P q · k k " kxkq # 1 2 q q Jk·kq (x)y 1 kxkq · = exp( q ) y 1 x . x 2 q q · " − 1+ q Jk·kq (x)x · # k k kxkq · Hence we have provided a second counterexample to Saint Raymond’s question. 8 C. Feng and L. Li

Remark 4.2. We remark that the assumption of q being an even integer in Theorem 4.1 can be replaced with q 3 being an odd integer. In the following we only establish the special case ≥ of q = 5, and leave the general situation as an exercise. Note t t 5 is a C1 function on R 7→ | | with derivative 5 t t3 (see also [5, §2.6], [7, p. 264] or [18, §2.6])1. Thus for any x, h l5, | | ∈ ∞ 5 5 5 5 x + h x = xk + hk xk k k5 − k k5 | | − | | Xk=1 h i ∞ 1 3 = 5 xk + thk (xk + thk) hkdt 0 | | Xk=1 Z ∞ ∞ 1 3 3 3 = 5 xk xkhk + 5 xk (xk + thk) xk hkdt | | 0 | | − Xk=1 Xk=1 Z h i ∞ 1 3 + 5 xk + thk xk (xk + thk) hkdt 0 | | − | | k=1 Z h i ∞X 3 =: 5 xk x hk + A + B. | | k k X=1 The main term, as a map of h, is a linear functional on (l5, ) with operator norm 5 x 4. k · k5 k k5 For the first remainder, it follows from H¨older’s inequality that ∞ 3 2 2 3 4 3 2 2 3 4 A 5 3 xk h + 3x hk + xk h 5 3 x h + 3 x h + x h . | |≤ | | k k| | | | k ≤ k k5k k5 k k5k k5 k k5k k5 Xk=1 h i h i For the second remainder, it follows from H¨older’s and Minkowski’s inequalities that ∞ 3 2 3 2 B 5 ( xk + hk ) h 5( x + h ) h . | |≤ | | | | k ≤ k k5 k k5 k k5 Xk=1 Therefore, G : x x 5 is everywhere differentiable on (l5, ) with derivative given by the 7→ k k5 k · k5 main term. For any x,y,h l5, ∈ ∞ 3 3 (JG(x) JG(y))h = 5 xk x yk y hk − | | k − | | k k=1 h i X∞ ∞ 3 3 3 3 = 5 xk x yk x hk + 5 yk x yk y hk | | k − | | k | | k − | | k k=1 h i k=1 h i X∞ ∞X 3 2 2 = 5 xk yk x hk + 5 xk yk x + xkyk + y yk hk, | | − | | k − · k k · | | k=1 k=1 X h i X

which, followed by generalized H¨older’s inequality, implies that

3 2 2 JG(x) JG(y) 5 x y x + 5 x y ( x + x y + y ) y . k − k≤ k − k5k k5 k − k5 k k5 k k5k k5 k k5 k k5 1 5 Consequently, G is a C function on (l , 5). Finally, one can deduce from (4.1) that G is 1 p k · k also a C function on (l , p) whenever p [1, 5]. k · k ∈ 1Lieb and Loss [18, §2.6] would probably write this derivative as 5|t|3t, while we prefer 5|t|t3. Differentiable maps with isolated critical points 9

5 Third example

Consider ∞ N l = x = (x1,x2, ) R : sup xk < · · · ∈ k | | ∞ n o with uniform norm x ∞ := sup xk . Obviously, k k k | | ∞ 2 1 x 2 x = k ||| ||| 2k k  X=1  is another norm on l∞ that is strictly weaker than . For any x, h l∞, k · k∞ ∈ ∞ ∞ 2 2 2 xk hk x + h x = hk + ||| ||| − ||| ||| 2k−1 2k Xk=1 Xk=1 =: V + V I. Note that ∞ xk J(x) : h hk 7→ 2k−1 k X=1 is a bounded linear functional on (l∞, ) with operator norm k · k∞ ∞ xk sup J(x)h = | | 2 x , | | 2k−1 ≤ k k∞ ||h||∞=1 k X=1 and ∞ h 2 VI k k∞ = h 2 . | |≤ 2k k k∞ k X=1 Thus x x 2 is an everywhere differentiable function on (l∞, ) with derivative at x 7→ ||| ||| k · k∞ given by J(x). For any x, z l∞, ∈ J(x) J(z) = sup (J(x) J(z))h = sup J(x z)h 2 x z ∞, k − k ||h||∞=1 | − | ||h||∞=1 | − |≤ k − k which implies that x x 2 is a C1 function on (l∞, ). Similar to the discussions in 7→ ||| ||| k · k∞ Sections 2 and 3 as well as considering Remark 3.3, it is not difficult to show that 1 x exp x (5.1) ∞ x2 7→ − k ! · 2k k=1 P on (l∞, ) is a non-open C1 map with a unique critical point at the origin. Hence we k · k∞ have provided a third counterexample to Saint Raymond’s question. Remark 5.1. The closed subset N c0 = x = (x1,x2, ) R : lim xk = 0 · · · ∈ k→∞ ∞ n o of (l , ∞), endowed with induced norm, is a Banach space. Similar to the above example, k · k 1 one can show that the map defined by (5.1) on c0 is a non-open C map with a unique critical point at the origin. 10 C. Feng and L. Li

6 Fourth example

Let (Ω, , P ) be a probability space, and denote for any p [1, ], F ∈ ∞ Lp = Lp(Ω, , P ) := f real-valued measurable on Ω : f pdP < (1 p< ) F { | | ∞} ≤ ∞ ZΩ 1 p with norm f p := ( f dP ) p , and || || Ω | | L∞ = L∞(ΩR, , P ) := f essentially bounded real-valued measurable on Ω F { } with standard uniform norm . According to H¨older’s inequality, k · k∞

f p1 f p2 (1 p

fhdP f p h p′ f p h p, Ω ≤ || || · || || ≤ || || · || || Z so h Lp 2 fhdP ∈ 7→ ZΩ p is a bounded linear functional on (L , p) with operator norm bounded above by 2 f p. k · k k k On the other hand, by (6.1), we have

2 2 h dP h p. Ω ≤ || || Z p p Therefore, G is everywhere differentiable on (L , p) with derivative at f L given by k · k ∈

p JG(f)h = 2 fhdP (h L ). ∈ ZΩ For any f, g Lp, ∈

JG(f) JG(g) = sup (JG(f) JG(g))h = sup JG(f g)h 2 f g p, k − k khkp=1 | − | khkp=1 | − |≤ k − k Differentiable maps with isolated critical points 11

1 p which implies that G is a C function on (L , p). Similar to the discussions in Sections 2 k · k and 3 as well as considering Remark 3.3, it is not difficult to show that 1 f exp f (6.2) 7→ − f 2dP ·  Ω  p 1 on (L , p) is a non-open C map with a uniqueR critical point at the origin. So we have k · k provided a fourth counterexample to Saint Raymond’s question. Remark 6.1. Let P be the Lebesgue measure on the σ-field of Borel measurable subsets of F Ω := [0, 1]. The family C(Ω) of all real-valued continuous functions on Ω is a closed subset of L∞(Ω, , P ), hence endowed with induced uniform norm, it is a Banach space. Similar to the F above example, one can show that the map defined by (6.2) on C(Ω) is a non-open C1 map with a unique critical point at the origin.

7 Weakly separable spaces

Having studied various examples in the previous four sections, we are now able to deal with some general Banach spaces by introducing a concept called weakly separable space. Definition 7.1. An infinite dimensional real Banach space X = (X, ) is said to be weakly k∞ · k separable if there exists a sequence of continuous linear functions lk on X such that { }k=1 x = 0 in X lk(x) = 0 for all k N. (7.1) ⇐⇒ ∈ Remark 7.2. (7.1) means x = y in X if and only if lk(x)= lk(y) for all k N, or equivalently ∈ ∞ x X : lk(x) = 0 = 0 . ∈ { } k=1 \  We further remark that it is of no harm to assume lk X∗ = 1 for all k N in (7.1). k k ∈ p Example 7.3. Let p [1, ], and define lk(x) = xk for x = (x1,x2, ) l and k N. ∈ ∞ p · · · ∈ ∈ Then it is easy to verify that (l , p) is weakly separable. In exactly the same way, one can k · k show that c0 (see Remark 5.1) is also weakly separable. Example 7.4. Let p [1, ], and let Lp(Rn) denote the p-times real-valued Lebesgue inte- ∈ ∞ grable functions on Rn with standard p-norm. To be clear, if p = then p-times and p-norm ∞ are understood as essentially bounded and uniform norm, respectively. We claim that Lp(Rn) ∞ is weakly separable. To prove this claim, we first pick all real-valued polynomials gk with { }k=1 rational coefficients on [0, 1]n, then define

lm1,...,m ,k : f f(y ,...,yn)gk(y m ,...,yn mn)dy dyn n 7→ 1 1 − 1 − 1 · · · Z[m1,m1+1]×···×[mn,mn+1] p Rn Zn N N on L ( ) for all (m1,...,mn, k) . If lm1,...,mn,k(f) = 0 for all k and some fixed n ∈ × ∈ (m ,...,mn) Z , then one can suitably apply the Stone-Weierstrass theorem [1, 15] to get 1 ∈

f(y ,...,yn)h(y ,...,yn)dy dyn = 0 1 1 1 · · · Z[m1,m1+1]×···×[mn,mn+1] for all real-valued continuous functions h on [m + 1,m ] [mn,mn + 1], which implies 1 1 ×···× that f vanishes almost everywhere on the same n-cube. This suffices to establish the claim. 12 C. Feng and L. Li

Example 7.5. In accordance with Remarks 5.1 and 6.1, it is straightforward to show that any infinite dimensional closed subspace of a weakly separable Banach space is also weakly separable (see also Remark 7.2).

Theorem 7.6. Any infinite dimensional separable real Banach space is weakly separable.

∞ Proof. Let xk be a sequence of non-zero elements of an infinite dimensional separable { }k=1 real Banach space (X, ) such that its closure is X. Applying the Hahn-Banach extension k · k theorem to λxk λ xk (λ R) on the one-dimensional linear subspace spanned by xk for 7→ k k ∈ ∗ ∞ each k, we see that there exists a sequence of continuous linear functionals lk X with { ∈ }k=1 lk X∗ = 1, such that lk(xk) = xk for all k. Then for any x = 0, by choosing a positive k k kxkk k 6 integer j such that xj x , one gets k − k≤ 4 x x lj(x)= lj(xj) lj(xj x) xj xj x x 2 xj x x k k = k k > 0. − − ≥ k k − k − k ≥ k k− k − k ≥ k k− 2 2 ∗ ∞ Thus (7.1) holds for lk X . This proves that (X, ) is weakly separable. { ∈ }k=1 k · k Theorem 7.7. The dual space of an infinite dimensional separable real Banach space is weakly separable.

∞ Proof. Let xk be a sequence of non-zero elements of an infinite dimensional separable { }k=1 real Banach space (X, ) such that its closure is X. For each k N, k · k ∈ ∗ lk : f X f(xk) R ∈ 7→ ∈ ∗ ∗ is a continuous linear functional on X . Suppose lk(f) = 0 for some f X and all k N, or ∈ ∞ ∈ equivalently f(xk) = 0 for all k N. Since f is continuous on X and xk is dense in X, ∈ { }k=1 one immediately gets f =0 in X∗. This proves that X∗ is weakly separable.

Remark 7.8. Theorem 7.7 includes l∞ and L∞(Rn) as typical examples, so weakly separable Banach spaces are not necessarily separable.

Our next result is in essence the same as the example given in Section 5.

Theorem 7.9. Let (X, ) be a weakly separable Banach space such that (7.1) holds for ∗ k∞ · k some sequence lk X with lk X∗ = 1 for all k N. Then { ∈ }k=1 k k ∈ 1 x exp x ∞ 2 7→ − lk(x) ! · 2k k=1 P on (X, ) is a non-open C1 map with a unique critical point at the origin. k · k

Proof. Since lk X∗ = 1 for all k N, k k ∈ ∞ 2 1 l (x) 2 x := k ||| ||| 2k  Xk=1  Differentiable maps with isolated critical points 13 is another norm on X and x x for all x X. For each q N, ||| ||| ≤ k k ∈ ∈ q

Xq := x X : lk(x) = 0 ∈ k=1 \  is a closed subspace of (X, ) with codimension q. Since X is infinite dimensional, one k · k ≤ can pick a non-zero element xq of Xq for each q N. Note then ∈ ∞ 2 1 lk(xq) 2 xq xq = k k (q N), ||| ||| 2k ≤ √2q ∈  k=Xq+1  which implies that is strictly weaker than on X. For any x, h X, ||| · ||| k · k ∈ ∞ l (x)l (h) x + h 2 x 2 = k k + h 2, ||| ||| − ||| ||| 2k−1 ||| ||| Xk=1 from which one can easily deduce that G : x x 2 on (X, ) is a differentiable function 7→ ||| ||| k · k with derivative at x X given by ∈ ∞ lk(x)lk(h) JG(x)h = (h X). 2k−1 ∈ Xk=1 For any x,y X, ∈ ∞ x y JG(x) JG(y) = sup (JG(x) JG(y))h = sup JG(x y)h k − k = 2 x y , k − k | − | | − |≤ 2k−1 k − k khk=1 khk=1 k X=1 which implies that G is a C1 function on (X, ). By composition rule, k · k 1 F : x exp( ) x 7→ −G(x) · on (X, ) is a C1 map with derivative at x X given by k · k ∈ 1 1 1 JF (x)h = exp( ) JG(x)h x + exp( ) h (h X). (7.2) −G(x) · G(x)2 · · −G(x) · ∈ 1 1 To be clear, the value of the function t [0, ) exp( ) 2 at t = 0 is understood as 0. ∈ ∞ 7→ − t t According to the discussions in the beginning part of Section 2 or (7.2), the origin is a critical point of F . Assuming next x is an arbitrary non-zero element of X, one can check that the solution h X, to JF (x)h = y, y X, is uniquely given by ∈ ∈ 1 2 JG(x)y 1 G(x) · h = exp( ) y 1 x . G(x) · − 1+ 2 JG(x)x · " G(x) · # So by Banach’s isomorphism theorem, every non-zero element of X is a regular point of F . Recall that the non-openness of F is guaranteed by Theorem 2.2. This finishes the proof of Theorem 7.9.

Considering Remark 7.2 and Theorem 7.9, we see that C1 maps with isolated critical points on weakly separable Banach spaces are not necessarily open. 14 C. Feng and L. Li

8 Further remarks

Remark 8.1. In sharp contrast to the infinite dimensional scenario, differentiable maps with isolated critical points do have nice properties in finite dimensional spaces. Apart from the open mapping property mentioned in the Introduction, differentiable vector fields with isolated critical points on Euclidean spaces are local homeomorphisms provided the dimension of the ambient space is higher than two [3, 17]. Question 8.2. Our general approach depends crucially on the fact that x x s is a non- 7→ ||| ||| negative real-valued function on (X, ) so that x exp( x −s) plays like a “black hole” k · k 7→ −||| ||| near the origin. Can anyone provide a counterexample to Saint Raymond’s question in a class of or some particular infinite dimensional complex Banach spaces?

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