# A (non trivial) continuous map on a Banach space which is nowhere Frechet differentiable

Assume that $X$ is a Banach space. Is there a continuous map $f:X\to X$ such that $f$ is nowhere Frechet differentiable, but its restriction to every finite dimensional subspace is every where Frechet differentiable?

## 1 Answer

Fr\'echet differentiability implies G\^ateaux differentiability, but the converse is true only for finite-dimensional Banach spaces, in general. As an example, the mapping $f:L^1[0,\pi]\rightarrow\mathbb{R}$ defined by $f(x)=\int_0^\pi\mbox{sin}(x(t))dt$ is every where G^ateaux differentiable, but nowhere Fr\'echet differentiable [9]

[9] M. Sova, Conditions for differentiability in linear topological spaces, Czechoslovak Math. J. (Russian) 16 (1966) 339­-362.

• I think in finite (2) dimension there is a Gateaux differentiable function whih is not Frechet differentiable. so in finite dimension, these two are not equivalent. :en.wikipedia.org/wiki/Fr%C3%A9chet_derivative – Ali Taghavi Mar 25 '14 at 13:21
• In the context of the paper the first statement was probably only for differentiation of norms. It seems to me that this does not affect this specific example. – user46855 Mar 25 '14 at 14:48
• Gateaux differentiability implies Frechet differentiability for Lipschitz mappings on finite dimensional spaces. – Bill Johnson Mar 25 '14 at 14:52
• Notice that non separability of $L_1^*$ is essentially for the example user46855 gave. Preiss' differentiability theorem says that every real valued Lipschitz function on a space with separable dual has a point of Frechet differentiability. – Bill Johnson Mar 25 '14 at 14:57
• @BillJohnson Prof. Johnson Thank you for your 2 interesting comments – Ali Taghavi Mar 25 '14 at 21:14