# For which values of $1 ≤ p ≤ \infty$ is $\{f_n\} \subset (C[0,1],\lVert\cdot\rVert_p)$ a Cauchy sequence?

The question is:

Let $\{f_n\} \subset (C[0,1], \lVert\cdot\rVert_p)$, where $f_n(x)=1-nx$, if $0\leq x \leq \frac1n$, and $0$ otherwise. For which values of $p$, $1 \leq p \leq \infty$, is $\{f_n\}$ a Cauchy sequence? When it is a Cauchy sequence, does it converge?

So, I know the definition of a Cauchy sequence. You have to start by letting $\epsilon > 0$, then there exists an $N \in \mathbb{N}$, such that if $n > m \geq N$, then $\lVert f_n - f_m \rVert_p < \epsilon$. But I can't even graph the function described above. Also, I don't know how a cauchy sequence can be related to the $\lVert \cdot \rVert_p$ .

If anyone could shed some light on this question, I would deeply appreciate it.

• The graph of the function you are looking at is the line connecting $(0,1)$ to $(1/n,0)$ and then turning into the horizontal line $(x,0)$. Nov 19, 2016 at 15:39

Note that $f_n(x)=0$ if $x≥1/n$ and $|f_n(x)|≤1$ if $x\in[0,1/n]$. For that reason:
$$\int_0^1|f_n(x)|^p\,dx≤\int_0^{1/n}1\,dx=1/n$$
So $$\|f_n\|_p≤\sqrt[\leftroot{-13}\uproot{2}p\quad ]{1/n}\to0$$ for all $p<\infty$.