Problem
Consider the space $C[-1,1]$, together with the norm defined by $\|f\|_1 = \int_{-1}^1|f|d\lambda$ (where $\lambda$ is the Lebesgue measure). For each $n$ define a function $f_n:[-1,1]\to\mathbb{R}$ by
\begin{align*}
f_n(x)=
\begin{cases}
0\quad &\text{if $-1\leq x\leq 0$},\\
nx\quad &\text{if $0<x\leq\frac{1}{n}$},\\
1\quad &\text{if $\frac{1}{n}<x\leq1$}.
\end{cases}
\end{align*}
Prove that $\{f_n\}$ is a Cauchy sequence in $C[-1,1]$. Then show there is now continuous function $f$ such that $\lim_{n\to\infty}\|f_n-f\|_1=0$.
My Attempt So Far
- Here is my attempt so far to prove the sequence $\{f_n\}$ is Cauchy:
Let $\epsilon$ be any positive number. We want to show that there is a positive integer $N$ such that $d(f_m,f_n)<\epsilon$ whenever $m\geq N$ and $n\geq N$. Write
\begin{align}
d(f_m,f_n) &= \|f_m-f_n\|_1\\
&= \int_{-1}^1|f_m-f_n|d\lambda\\
&= \int|f_m-f_n|\chi_{[a,b]}d\lambda\\
&= \sup\left\{\int hd\lambda:h\in\mathscr{S}_+\ \text{and}\ h\leq|f_m-f_n|\chi_{[a,b]}\right\}\\
&\dots
\end{align}
I got stuck here, don't know how to proceed next.
- For the nonexistence of a continuous function, I drew a diagram. I can see that the function would eventually goes to
\begin{align*}
f(x) =
\begin{cases}
0\quad &\text{if $-1\leq x\leq0$},\\
1\quad &\text{if $0<x\leq1$}.
\end{cases}
\end{align*}
which is clearly not continuous. But I don't know how to prove it rigorously.
Could someone please help me out? Thanks a lot in advance!
Best Answer
It is false that "Since $[-1,1]$ is compact, the $L^{1}$-norm is equivalent to $L^{\infty}$-norm."
The following is a famous counter-example (in probability theory). (one may argue that the given $f_n$ are not continuous, but we can always smooth out the function at the edge.)
Let $f_{1}=1_{[0,1]}$, $f_{2}=1_{[0,\frac{1}{2}]}$, $f_{3}=1_{[\frac{1}{2},1]}$, $f_{4}=1_{[0,\frac{1}{4}]}$, $f_{5}=1_{[\frac{1}{4},\frac{2}{4}]}$, $f_{6}=1_{[\frac{3}{4},\frac{4}{4}]},$$f_{7}=1_{[0,\frac{1}{8}]}$, $\ldots$.
Clearly $||f_{n}||_{1}\rightarrow0$ but $||f_{n}||_{\infty}=1$ for all $n$.
The correct way to solve the problem:
Firstly, we show that $(f_{n})$ is a Cauchy sequence with respect to the $L^{1}$-norm. Let $m,n\in\mathbb{N}$ with $m<n$. By direct calculation, we have that $$ |f_{n}(x)-f_{m}(x)|\leq2\cdot1_{[0,\frac{1}{m}]}(x). $$ That is, $|f_{n}(x)-f_{m}(x)|$ vanishes if $x\in[-1,0]\cup[\frac{1}{m},1]$ and $|f_{n}(x)-f_{m}(x)|\leq2$ if $x\in[0,\frac{1}{m}]$. Therefore, $\int_{-1}^{1}|f_{n}-f_{m}|\,d\lambda\leq\frac{2}{m}$. Hence, $(f_{n})$ is a Cauchy sequence with respect to the $||\cdot||_{1}$-norm.
Next, we show that $(f_{n})$ does not converges to any element in $C[-1,1]$. Prove by contradiction. Suppose there exists $f\in C[-1,1]$ such that $\int_{-1}^{1}|f_{n}-f|\,d\lambda\rightarrow0$. By Vitali Theorem, there exists a subsequence $(f_{n_{k}})$ and a measurable set $A\subseteq[-1,1]$ with $\lambda([-1,1]\setminus A)=0$ such that for each $x\in A$, $|f_{n_{k}}(x)-f(x)|\rightarrow0$ as $k\rightarrow\infty$. (i.e., $f_{n_{k}}-f\rightarrow0$ pointwisely a.e.).
For $x\in A$, clearly, $\lim_{k}f_{n_{k}}(x)=1_{(0,1]}(x)$. Hence, $f(x)=1_{(0,1]}(x)$ for $x\in A$. Since $f$ is continuous and $\lambda([-1,1]\setminus A)=0$, it follows that $f(x)=0$ for $x\in[-1,1]\setminus A$ too. (For, $[-1,1]\setminus A$ has measure zero $\Rightarrow$ it cannot contain any interior point $\Rightarrow$ for each $x\in[-1,1]\setminus A$, there exists a sequence $(x_{n})$ in $A$ such that $x_{n}\rightarrow x$. By the continuity of $f$...).
That is, $f=1_{(0,1]}$, which is a contradiction.