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Let $f\in L^P(\mathbb{R}^n)$. Define for all $n\in \mathbb{N}$ the truncator operator $T_n(x)$ as

$$ T_n(x)=\begin{cases} x, & |x|\le n,\\ \frac{xn}{|x|},&|x|>n.\end{cases} $$ If we denote $B(0,n)$ as the ball centered in $0$ and radius equal to $n$, and $f_n:=1_{B(0,n)}T_nf$, clearly the function is bounded by $1$. I want to apply the dominated convergence theorem for proving that $f_n\to f$ strongly in $L^p(\mathbb{R}^n)$. However, I can not see if $f_n(x)\to f(x)$ a.e on $\mathbb{R}^n$. Is this fact true and how can I prove it?. Please any help will be appreciated

Diego Vargas
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1 Answers1

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Let's check first that $f_n(x)\to f(x)$ almost everywhere. Indeed, fix $x\in\mathbb{R}^n$. Certainly, we find $n$ large enough such that $x\in B(0,n)$ and $n\geq |f(x)|$. In particular, it follows that $f_n(x) = f(x)$ for this and all larger $n$. Hence $f_n(x) \to f(x)$ almost everywhere.
Now notice further that $|f_n(x)| \leq |f(x)|$. Indeed, for $|f(x)|\leq n$, we have $T_n(f(x)) = f(x)$, hence $|f_n(x)|\leq |f(x)|$ (since the indicator $1_{B(0,n)}$ is $\leq 1$). For $|f(x)| > n$, we have that $|T_n(f(x))| = n < |f(x)|$, so $|f_n(x)|\leq |f(x)|$ as well.
Hence $f_n$ is dominated by $f\in L^p$ and you can apply the dominated convergence theorem.

Tobius
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