Showing that the space $C[0,1]$ with the $L_1$ norm is incomplete

Question

Can anyone think of a relatively easy counter example to remember, which demonstrates that the space $C[0,1]$ with the $L_1$ norm is incomplete?

Thanks!

Answer 1

This example works to show $C[0,1]$ is not complete with respect to the $L^p$ norm for all $1\leq p < \infty.$

Consider the piecewise linear function described by $$f_n(x)=\begin{cases} 1,&\text{if }0\le x\le\frac12\\ 1-n(x-\frac{1}{2}),&\text{if }\frac12\le x\le \frac{1}{2}+\frac{1}{n}\\\\ 0,&\text{if }\frac{1}{2}+\frac{1}{n}\le x\le 1\;. \end{cases}$$

Then $$ \| f_n - f_m \|_p = \left( \int^{1/2+1/n}_{1/2} |f_n(x)-f_m(x)|^p dx\right)^{1/p} \leq \left( \frac{1}{n} \right)^{1/p} \to 0$$

so $f_n$ is Cauchy. Suppose $f_n$ has limit $f\in C[0,1].$ Then $$\int^{1/2}_0 |f(x)-f_n(x)|^p dx \leq \|f-f_n\|_p^p \to 0$$ so $f(x)=1$ on $[0,1/2].$ Similarly we see $f(x) = 0$ on $[1/2,1]$, which is a contradiction.

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Of course, the calculations are easy to verify once you have the example, and no one actually remembers the explicit equations for such a thing. All you have to remember is that it is $1$ on $[0,1/2]$ then goes down very quickly to $0.$

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Answer 2

It's worth remembering that the continuous functions are dense in $L^1[0,1]$, so you can limit to any discontinuous function in $L^1[0,1]$ you desire. For a proof, see here.

For an explicit example, consider the sequence of functions defined for $x\in[0,1]$ by

$$f_n(x)=\begin{cases} \left(x+\frac{1}{2}\right)^n,&\text{if }0\le x<\frac{1}{2}\\\\ 1,&\text{if } \frac{1}{2}\le x\le 1\;. \end{cases}$$

One easily checks that these are continuous, but the sequence limits to a function $f$ defined by

$$f(x)=\begin{cases} 0,&\text{if }0\le x<\frac{1}{2}\\\\ 1,&\text{if } \frac{1}{2}\le x\le 1\;. \end{cases}$$

Answer 3

Just pick a sequence $\langle a_n:n\in\Bbb N\rangle$ in $\left(\frac12,1\right]$ converging to $\frac12$ and for $n\in\Bbb N$ let

$$f_n(x)=\begin{cases} 1,&\text{if }0\le x\le\frac12\\ 1-\frac1{a_n-1/2}\left(x-\frac12\right),&\text{if }\frac12\le x\le a_n\\\\ 0,&\text{if }a_n\le x\le 1\;. \end{cases}$$

It’s easily verified that $\langle f_n:n\in\Bbb N\rangle$ is Cauchy in $L^1[0,1]$, but its pointwise limit is the indicator function of $\left[0,\frac12\right]$, which is not $L^1$-equivalent to any continuous function on $[0,1]$.

Answer 4

One can try to remember $f_n(x)=\dfrac1{\mathrm e^{n(2x-1)}+1}$, which converges in $L^1$ to $f(x)=\mathbf 1_{x\leqslant1/2}$.

Answer 5

For $f\in L^1[0,1]$ we always have $$K_n*f\to f$$ in $L^1$ where $K_n$ is the Fejér-kernel. Now, $f_n =K_n*f$ is a trigonometric polynomial which certainly belongs to $C[0,1]$.

Answer 6

Here is an easy example that shows that $C[0,1]$ with the $L^1$ norm is not complete. Let $$f_n(x) = \begin{cases} nx,& 0 \leq x \leq \frac{1}{n} \\ 1,& \frac{1}{n} \leq x \leq 1\end{cases}.$$

It should not be hard to verify that $f_n$ is a cauchy sequence with respect to the $L^1$ norm. Now if there were a function $f$ such that $f_n \rightarrow f$ in the $L^1$ metric, we would necessarily have that $f$ be $1$ for $0 < x \leq 1$ and $f(0) = 0$. But such a function cannot possibly be continuous so that $C[0,1]$ with the $L^1$ norm is not complete.