Show that $f_n\to f$ in the norm $L^1(\mathbb{R})$ for $f\in L^1(\mathbb{R})$.

Question

Let $f\in L^1(\mathbb{R})$. Define $$f_n(x)=\begin{cases} f(x) & \text{if }|x|\leq n\\0 & \text{otherwise}\end{cases}.$$ Show that $f_n\to f$ i.n.

This seems really obvious, so I'm not sure how to go about actually showing it. If $f_n(x)=f(x)$ for all $|x|\leq n$, then, as $n\to\infty$, $f_n\to f$, obviously. Unless I'm missing something.

The other thing is the "i.n.", convergent "in norm". Can anyone explain how to do that? Would it just be what I outlined above? Or something else?

Answer 1

Convergent "in norm" means $$ \int_{\Bbb{R}}|f_n-f|d\mu\to0 $$ as $n\to\infty$. Since $f\in L^1(\mathbb{R})$ $$ \lim_{n\to\infty}\int_{[-n,n]}|f|d\mu=\int_{\Bbb{R}}|f|d\mu<\infty $$ Given $\epsilon>0$, there is a $N$ such that for any $n>N$ $$ \int_{\Bbb{R}\setminus[-n,n]}|f|d\mu<\epsilon $$ So $$ \int_{\Bbb{R}}|f_n-f|d\mu=\int_{\Bbb{R}\setminus[-n,n]}|f|d\mu<\epsilon $$

Answer 2

Use the Lebesgue dominated convergence theorem. Define $f_n = f\chi_{[-n,n]}$, which is the same as your definition. Then $|f_n(x)| \le |f(x)|$ for all $x\in\mathbb{R}$. Therefore, $|f_n-f| \le 2|f|$ and $\lim_n|f_n(x)-f(x)|=0$ for all $x$. Therefore, by the Lebesgue dominated convergence theorem, $$ \lim_n \|f_n-f\| = \lim_n\int_{\mathbb{R}} |f_n-f|dx = 0. $$