Proof verification convergence of sequence in $L^1$

Question

I want to examine if this sequence converges in $L^1$ or point wise. $$a_n(t)=n\chi_{[0,1/n)}(t)$$ I think the sequence is not Cauchy. We have $$|a_{2n}-a_{n}|=\int| 2n\chi_{[0,1/2n)}(t)-n\chi_{[0,1/n)}(t)| ~d\lambda=n\int| (2\chi_{[0,1/2n)}(t)-\chi_{[0,1/n)}(t))| ~d\lambda$$ Since $\chi$ vanishes outside $[0,1/n)$ I can restrict the integral there $$\begin{align}=n\int_{[0,1/n)}| (2\chi_{[0,1/2n)}(t)-\chi_{[0,1/n)}(t))| ~d\lambda &=n\int_{[0,1/2n)}| (2\chi_{[0,1/2n)}(t)-\chi_{[0,1/n)}(t))| ~d\lambda \\&+n\int_{[1/2n,1/n)}| (2\chi_{[0,1/2n)}(t)-\chi_{[0,1/n)}(t))| ~d\lambda\\&=n\lambda([0,1/2n))+n\lambda([1/2n,1/n))\\&=\frac{1}{2}+\frac{1}{2}=1\end{align}$$

So the sequence is not cauchy and cannot converge in $L^p$

I am not sure what to do about point wise convergence point wise it converges almost everywhere to 0

Answer 1

Your argument prove that the sequence $\left(a_n\right)_{n\geqslant 1}$ is not Cauchy in $\mathbb L^1$ hence cannot converge.

For the pointwise convergence, fix $t\in (0,1)$. For which $n$'s do we have that $t\in (0,1/n)$? What can you conclude for $\left(a_n(t)\right)_{n\geqslant 1}$?

Answer 2

For pointwise convergence, just test the limit at an arbitrary $x \in \mathbb{R}$. It helps a lot to draw the graph of the first few $a_n$; what do they look like? This should tell you what the limit is almost everywhere. In particular, you should have two distinct cases for what $a_n(x)$ converges to, depending as whether $x = 0$ or not.