This is actually a rather simple question, and I must be missing something – any help is appreciated. Let's consider the groups $(C^1(\mathbb{R}), +)$ and $(C^0(\mathbb{R}), +)$ and define the homomorphism $D: C^1 \rightarrow C^0$ such that $D(f) = f' = \frac{d(f)}{dx}$.
What is the kernel of $D$? Well, if $D(f) = 0$, that means $f'(x) = 0$, $\forall x \in \mathbb{R}$ which implies that $f$ is a constant function. So $\ker(D) = \mathcal{C}$, where $\mathcal{C} = \{f \in C^1(\mathbb{R}) | f(x) = c, \forall x \in \mathbb{R}, c \in \mathbb{R}\}$. By the first isomorphism theorem, this then implies $\operatorname{Im}(D) \simeq C^1(\mathbb{R})/\mathcal{C}$.
So far, this seems ok. But consider the function $f(x) = \begin{cases}-\frac{x^2}{2} & \mbox{if } x \leq 0 \\ \frac{x^2}{2} & \mbox{if } x > 0\end{cases}$. It is a $C^1(\mathbb{R})$ function, and its derivative is the absolute value of $x$, which is not differentiable. So $f \in C^0(\mathbb{R}) \setminus C^1(\mathbb{R})$. But, according to the above, $D(f)$ is an equivalence class of $C^1(\mathbb{R})/\mathcal{C}$. And I don't think this can be, because $D(f)$ is not even in $C^1(\mathbb{R})$.
Where did I make mistakes? And why?
Thanks in advance for any help!
Best Answer
The equivalence class you are looking for is the set of all functions of the form $f(x)+\text{const}$, and these are $C^1$. Also, it is $f'(x)$ that is in $C^0\setminus C^1$.