For a function $f:X\subset\mathbf{U}\to\mathbf{V}$ where $X$ is an open subset of the Banach space $\mathbf{U},$ and $\mathbf{V}$ is another Banach space, define its tangent $Tf:X\times\mathbf{U}\to\mathbf{V}\times\mathbf{V}$ by $Tf(x,u)=(f(x),Df(x)\cdot u)$ where $Df(x)$ is the standard derivative of $f$ at $x.$
I want to deal with the following problem:
Let $\mathbf{U}$ be the Banach space of real sequences $(x_1,x_2,\dots)$ such that $|\sqrt{n^3}x_n|$ is bounded and set $\|x\|=\sup\{|\sqrt{n^3}x_n|:n\geqslant1\}.$ Define $f:\mathbf{U}\to\mathbf{U}$ by $f(x)_n=f_n(x_n),$ where $f_n:\mathbb{R}\to\mathbb{R}$ is a smooth convex function satisfying $f_n(y)=0$ if $y\leqslant1/n$ and $f_n(y)=y-2/n$ if $y\geqslant3/n.$ Show that given such an $f,$ the map $Tf$ exists and is continuous, $\|Df(x)\|$ is locally bounded, but $f$ is not $C^1.$
Supposedly this is an illustrative example of when the continuity of $Tf$ fails to be equivalent to $C^1$ when dealing with infinite-dimensional Banach spaces, due to L. Rosen. I am having trouble with the details of this proof. Is there a particularly intuitive way of proving this? Thanks.