If $u\in L^p(\Omega;\mathbb R^d)$, then $u_i\in L^p(\Omega)$ for all $i\in\left\{1,\ldots,d\right\}$. Does the reverse hold true?

Question

Let $\Omega\subseteq\mathbb R^d$ and $u:\Omega\to\mathbb R^d$ be Borel measurable. Since $$|u_i|\le\left\|u\right\|_2\;\;\;\text{for all }i\in\left\{1,\ldots,d\right\}$$ we obtain $$\left\|u_i\right\|_{L^p(\Omega)}\le\left\|u\right\|_{L^p(\Omega;\mathbb R^d)}\;\;\;\text{for all }i\in\left\{1,\ldots,d\right\}$$ for all $p\in [1,\infty)$.

So, if $u\in L^p(\Omega;\mathbb R^d)$, then $u_i\in L^p(\Omega)$ for all $i\in\left\{1,\ldots,d\right\}$. Does the reverse hold true?

Unfortunately, I was neither able to prove it nor to find a counterexample.

Answer 1

Since on $\mathbb{R}^n$ any two norms are equivalent there is a constant $C$ such that $$||v||_2\le C\sum_{i=1}^n|v_i|$$ for any $v=(v_1\dots,v_n)$. (so, yes).