Let f_n=1_(n,n+1) in L^1(ℝ) . Is it Cauchy? If so, does it have a limit in L^1 ?

It is not Cauchy, because \|f_n-f_m\|_1=2 for n≠ m

It is Cauchy and converges to 0

It is Cauchy and converges to 1_ℝ

Exercise · ZW-RIESZ-FISCHER-THEOREM-010 · ★★★ Advanced

Let \( f_n=\mathbf{1}_{(n,n+1)} \) in \( L^1(\mathbb{R}) \). Is it Cauchy? If so, does it have a limit in \( L^1 \)?

Choose one option.

Show answer

D. It is not Cauchy, because \( \|f_n-f_m\|_1=2 \) for \( n\neq m \)

Explanation

The intervals are disjoint, so the distance between two distinct terms is the sum of their measures, namely \( 2 \). Since it does not tend to zero, it is not Cauchy.

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Review the concept: Riesz–Fischer theorem

The Riesz-Fischer theorem states that the space \( L^p(\mu) \) is complete for \( 1 \le p \le \infty \), that is, every Cauchy sequence in it converges to an element of \( L^p(\mu) \).

\[\lim_{n,m\to\infty}\|f_n-f_m\|_p=0 \implies \exists f\in L^p:\ \|f_n-f\|_p\to 0\]

Let \( f_n=\mathbf{1}_{(n,n+1)} \) in \( L^1(\mathbb{R}) \). Is it Cauchy? If so, does it have a limit in \( L^1 \)?

Explanation

Review the concept: Riesz–Fischer theorem

More practice problems: riesz–fischer theorem