I'm considering the mapping $\Psi: C^2([0,1])$ to $C^1([0,1])$ via: $f(x) \mapsto f(x)+x\cdot f'(x)$.
Is this mapping a homeomorphism? It should be continuous given that, for any sequence $(f_n) \in C^2([0,1])$ such that $(f_n) \to f$, it is necessary for $C^2$ convergence that the functions, their first, and their second derivatives uniformly converge. Thus:
$$\|(f_n-f) +x\cdot (f_n'-f')\|_\infty \le \|(f_n-f) + (f_n'-f')\|_\infty \le \|f_n-f\|_\infty +\|f_n'-f'\|_\infty$$
where the right-hand side goes to zero, and an analogous argument holds for the derivative sequence.
But is it a homeomorphism? I'm having trouble with a sequential argument for the inverse, and feel like I may just be missing something. It seems intuitive that $\Psi$ should be onto due to solutions of first-order linear ODEs existing. Do things break with injectivity? Thanks for any guidance!
Consider the functions
$$h_n(x) = \begin{cases} n - n^2 x &, 0 \leqslant x \leqslant \frac{1}{n}\\ \quad 0 &, \frac{1}{n} \leqslant x \leqslant 1,\end{cases}$$
$g_n(x) = \int_0^x h_n(t)\,dt$ and $f_n(x) = \int_0^x g_n(t)\,dt$. By construction, $f_n \in C^2([0,1])$, and
$$\lVert f_n\rVert_{C^2([0,1])} \geqslant \lVert h_n\rVert_\infty = n.$$
But
$$\lVert\Psi(f_n)\rVert_{C^1([0,1])} = \lVert f_n + xg_n\rVert_\infty + \lVert 2g_n + xh_n\rVert_\infty \leqslant \lVert f_n\rVert_\infty + 3\lVert g_n\rVert_\infty + \lVert xh_n\rVert_\infty.$$
Now $0 \leqslant g_n(x) \leqslant \frac{1}{2}$, and therefore also $0 \leqslant f_n(x) \leqslant \frac{1}{2}$, so $\lVert f_n\rVert_\infty + 3\lVert g_n\rVert_\infty \leqslant 2$. And since $0 \leqslant h_n(x) \leqslant n$, and $h_n(x) = 0$ for $x \geqslant \frac{1}{n}$ we have $\lVert xh_n\rVert_\infty \leqslant \frac{1}{n}\cdot n = 1$, so altogether
$$\lVert\Psi(f_n)\rVert_{C^1([0,1])} \leqslant 3.$$
Thus
$$\inf \bigl\{ \lVert \Psi(f)\rVert_{C^1([0,1])} : \lVert f\rVert_{C^2([0,1])} = 1\} = 0$$
and $\Psi$ has no bounded inverse (defined on the range) of $\Psi$. By the open mapping theorem (since $\Psi$ is injective), $\Psi$ is not surjective.