If you have a truncation $T_{k}u$ defined as:
$$ T_{k}u := \begin{cases} u,& \text{ if }~ |u(x)| \leq 1\\ k\frac{u}{|u(x)|}, & \text{ if }~|u(x)| > k \end{cases} $$
If you consider the truncation of a function $u \in L^{p}(\Omega)$, $p > 1$ and $\Omega$ is bounded. Then how would you show that $T_{k}u \rightarrow u$ in $L^{p}(\Omega)$. It is clear that $T_{k}u \rightarrow u$ pointwise a.e. and by the Dominated convergence Theorem you could show that $T_{k}u \rightarrow u$ in $L^{1}(\Omega)$. I'm finding it hard to show convergence of truncation in $L^{p}(\Omega)$.
Does anyone have any idea how this can be shown?
The Dominated Convergence Theorem works in $L^p$ spaces, too: if the dominating function is $L^p$ (with $1\le p < \infty$) then the convergence is in $L^p$ sense. To wit, assume that $$|f_n| \le g \in L^p,$$ with $p>1$ (the case $p=1$ being the DCT). Then the pointwise limit $f$ of $f_n$ (which exists by assumption) is dominated by $g$ and hence $f\in L^p$. Moreover, using the convexity inequality $$(a+b)^p\le 2^{p-1} (a^p+b^p), \qquad \forall a,b \ge 0$$ we see that $$ |f_n-f|^p\le C g^p, $$ and now the DCT gives convergence $f_n \to f$ in $L^p$ sense.