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

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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 $M > 0$.

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  • $\begingroup$ Actually, come to think of it, the Dyadic decomposition is a bit of a red herring. You can just define $f(t,x) = \chi_{t^{-1}}(x)$ for $t \neq 0$ and $f(0,x) = 0$ and get the same effect. $\endgroup$
    – Willie Wong
    Commented Feb 17, 2021 at 4:18
  • $\begingroup$ Thank you for the answer! I have question on the last equality. I have computed that $$\|f\|_{L^\infty([0,T],L^p)}=ess\sup_{t\in [0,T]}\left| t^{-\frac{N}{p}}\|\chi\|_p \right|$$ The last value seems depends on $p$ and $N$. I am not sure if I computed rightly. Could you check it out for me? $\endgroup$
    – Lev Bahn
    Commented Feb 17, 2021 at 4:44
  • $\begingroup$ The form of $\chi_s$ is chosen so that $\|\chi_s\|_{L^p(\mathbb{R}^N)} = \|\chi\|_{L^p(\mathbb{R}^N)}$. You probably did your change of variables wrong. $\endgroup$
    – Willie Wong
    Commented Feb 17, 2021 at 4:46
  • $\begingroup$ You are right! Thank you so much! Do you know by any chance the general reason why that limit hold for continuous function but not $L^\infty$ function? $\endgroup$
    – Lev Bahn
    Commented Feb 17, 2021 at 4:54
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    $\begingroup$ The idea is very similar to the basic analysis statement that a continuous function on a compact is uniformly continuous (and hence bounded), but a continuous function on a open can be unbounded. For example, if your underlying space is $(0,T)$ and not $[0,T]$, then the $f(t,x)$ I defined in the comments is actually in $C^0((0,T); L^p)$, but cannot be extended to $C^0([0,T];L^p)$. But that's as far as the intuitions go. $\endgroup$
    – Willie Wong
    Commented Feb 17, 2021 at 5:06

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