Splitting function from $L^{n/2}$

Question

Let $V\in L^{n/2}$, $n\geq 3$. I want to show that for every $\varepsilon>0$, there are $||V_{1}||_{L^{n/2}}\leq \varepsilon$ and $V_{2}\in L^{\infty}$ sucht that $$V=V_{1}+V_{2}$$

Answer 1

This holds in $L^p$ for any any $p \in [1,\infty).$ Indeed if $V \in L^p,$ then define, $$ V^{(M)}(x) = \min\{M,\max\{V(x),-M\}\}. $$ Note that $\lVert V^{(M)}\rVert_{L^{\infty}} \leq M$ for all $M,$ so each $V^{(M)} \in L^{\infty}.$ Since $V^{(M)} \rightarrow V$ almost everywhere with $\lVert V^{(M)} \rVert_{L^p} \leq \lVert V \rVert_{L^p},$ by the dominated convergence theorem (applied to $|V-V^{(M)}|^p)$ we have, $$ \lim_{M \rightarrow \infty} \lVert V - V^{(M)} \rVert_{L^p} = 0. $$ Hence given $\varepsilon > 0,$ we choose $M$ large enough so $\lVert V-V^{(M)} \rVert_{L^p} \leq \varepsilon$ and take $V_1 = V-V^{(M)},$ $V_2 = V^{(M)}.$