Intereting Posts

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continuous monotonic function

I’d like to show that $C([0,1])$ (that is, the set of functions $\{f:[0,1]\rightarrow \mathbb{R} \, \textrm{ and } \, f \, \textrm{is continuous} \}$ is not a complete mertric space under the $L_1$ distance function:

$$

d(f,g) = \int_0^1 |f(x)-g(x)|dx

$$

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- Showing that a limit exists and showing $f$ is not integrable.

I can find counter examples (for example, here) but would rather prove it using definitions and principles so that I do not have to rely on committing specific degenerate sequences to memory.

Since all compact metric spaces are complete, I have to figure that the place to start is to show that $C([0,1])$ is not compact and that somehow an infinite cover allows for a divergent Cauchy sequence. However, I don’t how to show this (or if it’s even the right approach to take).

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$C[0,1]$ can be embedded as a subspace of $L^1[0,1]$. It is dense in $L^1$, but not equal to $L^1$. Therefore it is not closed, and hence not complete.

Since all compact metric spaces are complete, I have to figure that the place to start is to show that $C([0,1])$ is not compact and that somehow an infinite cover allows for a divergent Cauchy sequence.

It is certainly not compact, but the logic is off here. Compact metric spaces are complete, but complete metric spaces need not be compact. For example, think of $\mathbb R$ with its usual metric, or any other nonzero complete normed space.

I will assume you want to show that $C[0,1]$ is not a complete subspace of $L^1$.

Here is a counterexample-free approach.

Suppose $C[0,1]$ is a complete subspace of $L^1[0,1]$, so that it is in particular closed. Since it is clearly a proper subspace, the Hahn-Banach theorem tells us that there is a non-zero continuous linear functional $\phi:L^1[0,1]\to\mathbb R$ such that $\phi$ vanishes on $C[0,1]$. Now the fact that the dual of $L^1[0,1]$ can be identified with $L^\infty[0,1]$ allows us to translate this: there exists a non-zero function $g\in L^\infty[0,1]$ such that $\int_0^1 fg=0$ for all $f\in C[0,1]$.

That this is not possible is a standard result in measure theory.

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