LetB and C be complex Banach spaces, and let F : B → C be a Fr´echet differentiable map satisfying kF (x)k ≤ kxk for all x ∈ B

Problem 11037

(American Mathematical Monthly, Vol.110, October 2003) Proposed by M. L. J. Hautus (Nederlands).

LetB and C be complex Banach spaces, and let F : B → C be a Fr´echet differentiable map satisfying kF (x)k ≤ kxk for all x ∈ B. Show that F is linear.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Since F is a Fr´echet differentiable map then for all x0 ∈ B there exists a bounded linear mapping Ax0 : B → C such that

x→xlim0

kF (x) − F (x⁰) − Ax0(x − x⁰)k

kx − x0k = 0.

This is actually a complex differential and it is well known that F behaves like a holomorphic function: for all Ψ ∈ C^′ and for all x ∈ B, the map

f (ζ) = Ψ(F (ζx))

is holomorphic in C. By the inequality kF (x)k ≤ kxk, we have that F (0) = 0 and f (0) = 0. Hence also the map

g(ζ) = f (ζ)

ζ = Ψ F (ζx) ζ



is holomorphic in C and

|g(ζ)| ≤ kΨk ·kF (ζx)k

|ζ| ≤ kΨk ·kζxk

|ζ| = kΨk · kxk,

that is g is an entire bounded function and, by Liouville theorem, it is constant. Since C^′ separates points in C then also the map F (ζx)/ζ is equal to a constant (which depends on x): F (ζx) = ζcx. The differentiability in x⁰= 0 yields

ζ→0lim

kF (ζx) − F (0) − A0(ζx)k

kζxk = lim

ζ→0

kζcx− ζA0(x)k

kζxk =kcx− A0xk kxk = 0.

Hence cx= A⁰x and F is linear: F (x) = A⁰x.