Let $K \subset L^p([0,1])$. Let $f^+(x):= \max\{0,f(x)\}$, for all $x \in [0,1]$, be non-negative part of $f(x)$. Show that the subset of $K$, denoted by $K^+ =\{f^+: f \in K\}$ is also compact.
I am presenting my attempt to prove this question, and looking for comments or verifying.
My attempt
Let $\{f^+_n\}$ be a sequence in $K^+$. Then, there exists $\{f_n\}$ in $K$ such that $f^+_n(x) = \max\{0,f(x)\}$ for all $x \in [0,1]$ and $n \in \mathbb{N}$. Since $K$ is compact, then there exists $\{f_{n_k}\}$ in $K$ and $f \in L^p ([0,1])$ such that $\|f_{n_k}-f\|_p \to 0$ as $k \to \infty$. Define $f^-(x) := \min \{ 0, f(x)\}$. Then we have $f= f^+ - f^-$ (I am not sure of this)
$$\|f^+_{n_k} - f^+\| \leq \|f_{n_k} -f \|_p \to 0 \,\,\, \text{ as } \,\,\, k \to \infty.$$