Is $T':L^2(\Omega) \to L^2(\Omega)$ continuous? Here, $k$ is a fixed number. Let $$T(x) = \begin{cases}
-k &x \in (-\infty, -k]\\
x &x \in (-k, k)\\
k &x \in [k, \infty)
\end{cases}.$$
So $$T'(x) = \begin{cases}
0 &x \in (-\infty, -k)\\
1 &x \in (-k, k)\\
0 &x \in (k, \infty)
\end{cases}.$$
Let $\Omega$ be a bounded domain. 
Problem 1
Suppose $f_n \to f$ in $L^2(\Omega)$. Does it follow that $T'(f_n) \to T'(f)$ in $L^2(\Omega)$ as $n \to \infty$?

Problem 2
Suppose $f_n \to f$ in $H^1(\Omega)$. Does it follow that $T(f_n) \to T(f)$ in $H^1(\Omega)$ as $n \to \infty$?
I can show continuity in $L^2$ using the DCT. For the $H^1$ seminorm, not sure how to find a dominant function on $|T'(f_n)\nabla f_n| \leq |\nabla f_n|$ independent of $n$. And as pointed out in the comment, $T'$ is not defined at $\pm k$.

 A: For problem 1, you have an issue that by your current definition, $T'(f)$ is undefined if $f$ takes on the value $\pm k$ (on a set of positive measure).  Indeed, if $f$ is the constant function $k$, then $T(f)$ is undefined everywhere.
You could try to fix this by setting $T'(k) = 0$ or $T'(k) = 1$ or something, but it won't get you continuity.  Indeed, it is not hard to prove that:

If $g : \mathbb{R} \to \mathbb{R}$ and $f \mapsto g \circ f$ is continuous on $L^2(\Omega)$, then $g$ is continuous.

(Hint: You only have to think about $f$ ranging over the constant functions.)
A: Problem 1: as already noted, this is not true. See Nate Eldredge's answer and also this thread.
Problem 2: this is true. The truncation function can be written as a sum of $\max$ and $\min$, which have nice properties (applying these functions to a differentiable function preserves differentiability, they are sequentially continuous etc). See Lemma 2.89 on page 35 in this book for a proof for a generalised truncation function.
