Constructing a sequence of functions, not Cauchy

Question

I'm working in the set $B = \{ f \in C[0,1] : \int_0^1 f(x)dx \leq 1\}$. I'm constructing an argument to show that there exists at least one sequence that has a subsequences satisfying the property that $d(f,g) = \int_0^1 \left| f(x) - g(x) \right| dx = k$ for some fixed $k$ and $\forall f,g$ in the sequence. I've been told it's quite obvious to think of an example, but I can't seem to come up with it.

Answer 1

Assuming we can take $k\leq 1$, we can define such a sequence $\{f_n\}$ by letting $f_n$ have a sawtooth spike enclosing area $k/2$ on the interval $[2^{-n}, 2^{-n-1}]$ (for example, an isosceles triangle of height $2^{n+1}k$ based on this interval), and letting it be $0$ everywhere else.

It is easy to see that each of these are continuous, and they all clearly satisfy $\int_0^1 f_n = k/2<1$. Moreover, given distinct $f_n$ and $f_m$ in the sequence, these define spikes on disjoint intervals, so there is no cancellation when we form their difference. Indeed, $|f_n-f_m|$ has spikes enclosing positive area $k/2$ on the intervals $[2^{-n}, 2^{-n-1}]$ and $[2^{-m}, 2^{-m-1}]$ and is zero elsewhere, so $\int_0^1 |f_n-f_m| = k/2+k/2=k$.

It's worth noting that the existence of such a sequence proves that the closed unit ball in $C[0,1]$ with $L^1$ norm is not compact, since no subsequence of $\{f_n\}$ converges.

Answer 2

Let the elements of a sequence $(f_n)_{n=1}^{\infty}$ be defined by $$ f_n:=n(n+1)\chi_{[\frac 1{n+1},\frac 1{n}]}\ , $$ where $\chi_A$ is the characteristic function of a set $A$. It is not hard to see that $\int_0^1 f_n(x)dx = 1$ for all $n\in \Bbb N$. Since for any two distinct $f_n$ and $f_m$, their supports intersect on a set of zero measure, we have $$\begin{align} d(f_n,f_m) &= \int_0^1 \left| f_n(x) - f_m(x) \right| dx \\ &= \int_0^1 \left| f_n(x) \right| dx + \int_0^1 \left|f_m(x) \right| dx \\ &= 2\ . \end{align}$$

Edit: As Mr.Glare has pointed out, I missed the condition that $f_n$ need to be continuous.