$\frac{d}{dt}$ on $\mathbb{R}$ is not a Fredholm operator? I encountered a statement in a book that $\frac{d}{dt} : L^2_1(\mathbb{R}) \longrightarrow L^2 (\mathbb{R})$ is not a Fredholm operator, where $L^2_1$ is the first Sobolev space of the $L^2$ space. Consider a sequence of function $g_n$ such that $g_n \equiv 1$ on $[-n,n]$, increasing on $[-n-1,-n]$ and decreasing on $ [n,n+1]$, and vanishes outside of $[-n-1,n+1]$. We have $\operatorname{lim}_{n\longrightarrow \infty}\limits \lVert g_n \rVert_{L^2_1} = \infty$ and $\lVert \frac{d}{dt} g_n \rVert_{L^2} \leq C$ for some constant $C$. Then the author says if $\frac{d}{dt} $ is Fredholm, then the sequence will imply a non-trivial kernel which is a contradiction. My question is how he knows the implication of non-trivial kernel?
Thank you
edit:
The book I'm reading is "Floer homology groups in Yang-Mills theory" by Donaldson and this problem is on page 58.
 A: In general, if $L : A \to B$ is Fredholm and injective, then $L : A \to L(A)$ is bounded bijective between Banach space (Since $L$ has closed range) and so has a bounded inverse, by the bounded inverse theorem. As a result, there is no sequence $g_n \in A$ so that 
$$ \|g_n\|_A \to \infty \ \ \text{and } \ \ \|Lg_n\|_B \le C$$
Since this would imply that the inverse is unbounded. As a result, $\frac{d}{dt} : L_1^2 \to L^2$ cannot be Fredholm since it is injective. 
Remark:
Indeed, it is easy to see that $\frac{d}{dt}$ is not Fredholm as it does not has closed range: For all $f\in C_c(\mathbb R) \subset L^2(\mathbb R)$, the function 
$$ f_n = f- \frac{c}{n} \chi_{[0,n]},\ \ \text{where } \ \ c = \int_{\mathbb R} f$$
satisfies $\|f-f_n\|_{L^2(\mathbb R)} \to 0$ and $f_n = \frac{d}{dt} g_n$, where 
$$g_n(x): = \int_{-\infty }^x f_n(s)ds \in C^{0,1}_c(\mathbb R).$$
Thus the image of $\frac{d}{dt}$ is dense in $L^2$ but is not the whole $L^2$ (pick an element $f \in C_c(\mathbb R)$ with $c\neq 0$ for example)
