Let me be more specific.


Let $T>0$ be a finite real number.


We know that if $f\in C([0,T],L^p(\mathbb{R}^N))$ is complex valued then the statement $\|1\_{\{|f|>M\}}|f|\|\_{L^\infty([0,T], L^p(\mathbb{R}^N))}\rightarrow 0$ as $M\rightarrow \infty$ is true by dominated convergence theorem with the dominant function $f$.


But is the same statement true for $f\in L^\infty([0,T],L^p(\mathbb{R}^N))$?


I thought it is just true with same argument. However, my teacher says it is not true. He does not let me know why it is not true.


Could I have the answer from here? Thanks in advance!


**Addition**: $f\in C([0,T],L^p(\mathbb{R}^N))$ means, for every $t\in [0,T]$ and every $\varepsilon>0$ there exists $\delta>0$ such that
$$\|f(t)-f(s)\|\_{L^p(\mathbb{R}^N)}<\varepsilon$$
for every $s\in [0,T]$ such that $|t-s|<\delta$.


**Addition2** : The symbol $\{|f|>M\}$ means $\{(t,x)\in [0,T]\times \mathbb{R}^N : |f(t,x)|>M\}$



$\|1_{\{f>M\}}f\|_{L^\infty_t L^p}\rightarrow 0$ as $M\rightarrow \infty$ for $f\in L^\infty([0,T],L^p)$?
Fix $\chi$ an $L^p$ function.
Let $\chi\_s(x) = s^{N/p} \chi(sx)$ for $s > 0$.


Define $f$ by considering its values on the dyadic interval $t\in [2^{k}, 2^{k+1})$. For
$$ f|\_{t\in [2^k, 2^{k+1})} = \chi\_{2^{-k}}(x) $$


Then you have $$\|f\|\_{L^\infty([0,T] ,L^p(\mathbb{R}^N))} = \|\chi\|\_{L^p} = \|1\_{\{|f|>M\}}f\|\_{L^\infty([0,T] ,L^p(\mathbb{R}^N))}$$
for every $M > 0$.


