Prove that $(f_n)$ is a Cauchy sequence, but that it does not converge relative to $N_1$ to an element of $C[0,1].$

Question

Let $C[0,1]$ be the linear space of continuous functions, and define a norm as $N_1 = \int |f|d\mu.$ (a Riemann integral). Let $$f_n(x) = \begin{cases} 0, & \text{if $0 \leq x \leq \frac{1}{2} - \frac{1}{2n}$} \\ \text{a linear function}, & \text{if $\frac{1}{2} - \frac{1}{2n} \leq x \leq \frac{1}{2}$} \\ 1, & \text{if $\frac{1}{2} \leq x$} \end{cases}$$ Prove that $(f_n)$ is a Cauchy sequence, but that it does not converge relative to $N_1$ to an element of $C[0,1].$

When I have a sequence of functions, like $f(x) = x + \frac{1}{n},$ it is very direct to take the limit. But in the case of the problem, the $n$ only lies in the domain. How do I interpret this?

As $\lim_{n \to \infty}{\frac{1}{2} - \frac{1}{2n}}= \frac{1}{2},$ I see that $f(x)$ can turn into $$f(x) = \begin{cases} 0, & \text{if $0 \leq x < \frac{1}{2}$} \\ 1, & \text{if $\frac{1}{2} \leq x$} \end{cases},$$ which is a non continuous function, so it is true that $f \notin C[0,1].$ And, as $f$ is a uniformly bounded function, by the dominated convergence theorem, $\int fd\mu = \lim_{n \to \infty}\int|f_n|d\mu$.

Because the $f_n$ are continuous, then $\lim_{n \to \infty}\int|f_n|d\mu = \int\lim_{n \to \infty}|f_n|d\mu,$ and the right hand side of the equality is what I interpreted as "taking the limit in the domain of $f_n.$" From this I concluded that $(f_n)$ converges to an element that is not in $C[0,1],$ with respect to $N_1.$ Is this a correct reasoning or not?

I am more stuck at proving that $(f_n)$ is uniformly Cauchy. On the left hand side of the function, near $f(\frac{1}{2} - \frac{1}{2n}),$ $f_n - f_m = 0,$ so the integral is $0,$ which is what I have to show. While on the RHS near $f(\frac{1}{2} - \frac{1}{2n})$ I see how the functions are moving, but don't know how to express it with Epsilons. I am not being able to apply the definition to prove this, and don't have a clue at what to do.

Answer 1

Take $m,n,p\in\mathbb N$ with $m,n\geqslant p$. Then both $f_m$ and $f_n$ are null outside $\left[\frac12-\frac1{2p},\frac12+\frac1{2p}\right]$ and both take values between $0$ and $1$ with that interval. So,$$N_1(f_m,f_n)=\int_{\frac12-\frac1{2p}}^{\frac12+\frac1{2p}}|f_m-f_n|\leqslant\frac2p.$$Therefore, given $\varepsilon>0$, you just take $p\in\mathbb N$ such that $\frac2p<\varepsilon$.

If $(f_n)_{n\in\mathbb N}$ converges to some $f\in C\bigl([0,1]\bigr)$, let $x_0\in[0,1]$. If $x_0<\frac12$, then $f(x_0)=0$. In fact, if $f(x_0)>0$, then $f(x)>0$ in some inteval $(x_0,-\delta,x_0+\delta)$, with $x_0+\delta<\frac12$. Take $n\in\mathbb N$ such that $\frac12-\frac1{2n}>x_0+\delta$. Then $N_1(f_n,f)\geqslant\int_{x_0-\delta}^{x_0+\delta}\bigl|f(x)\bigr|\,\mathrm dx>0$. This contradicts the fact that $(f_n)_{n\in\mathbb N}$ converges to $f$. By the same argument, $f(x)=1$ when $x>\frac12$. But there is no continuous function $f$ with $f(x)=0$ is $x<\frac12$ and $f(x)=1$ if $x>\frac12$.

Answer 2

Note that the limit in the $N_1$ sense need not be a pointwise limit. For example, if we had $f_n(x)=\max\{0,1-nx\}$ instead, then $f_n\to 0$ relative to $N_1$, but the pointwise limit is non-continuous. Hence one more step is needed to show that there is no $N_1$-limit of the sequence $\{f_n\}$ of the problem statement. But for that it suffices no remark that any continuous $f$ differs from the pointwise limit by so much as to guarantee positive $N_1$-distance from it.

That $\{f_n\}$ is a Cauchy sequence, is not that hard: If $n,m>N$, then $f_n$ and $f_m$ differ at most on the interval $[\frac12-\frac1{2N},\frac12]$ and differ by at most $1$ there, thus making $\int |f_n-f_m|\,\mathrm d\mu\le\frac1{2N}$