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I have a more general question about showing that a normed vector space is not complete with respect to a given norm. We know that $C([0,1])$ is complete with respect to the sup norm. To show that $C([0,1])$ is not complete with respect to $L^1$-norm, one can exhibit an $L^1$-Cauchy sequence $f_n \in C([0,1])$ and show that it cannot converge to a continuous function with respect to the $L^1$-norm. However, does it also suffice to show that the sup norm and $L^1$-norm are not equivalent in this space? Consider the sequence of functions $f_n(x) = x^n$. Then $$\|f_n\|_{\infty} = 1 \mbox{ for all } n \in \mathbb{N}$$ while $$\|f_n\|_{L^1} = \int_0^1 x^n dx = \frac{1}{n+1} \rightarrow 0 \mbox{ as } n \rightarrow \infty$$ so there does not exist a $C > 0$: $$\|f_n\|_{\infty} \leq C \|f_n\|_{L^1} \mbox{ for all } n \in \mathbb{N}.$$ Since these norms are not equivalent and $C([0,1])$ is complete with respect to the sup norm, does it necessarily follow that $C([0,1])$ is not complete with respect to the $L^1$-norm?

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  • $\begingroup$ This looks to be a very nice first question! Welcome to the site. Don't be a stranger! $\endgroup$ – The Count Dec 24 '16 at 20:18
  • $\begingroup$ Thank you! Have been a longtime lurker and would like to be more involved. $\endgroup$ – xk3 Dec 24 '16 at 20:19
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    $\begingroup$ See math.stackexchange.com/questions/6616/… where it is shown that in general, a vector space can have two inequivalent complete norms (assuming the axiom of choice). $\endgroup$ – Nate Eldredge Dec 24 '16 at 21:14
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No, that would not be a valid proof. As you have stated, two norms are equivalent iff they have the same Cauchy sequences and thus, if one norm is complete, any equivalent norm is also complete. However, two non-equivalent norms can both make a space complete. This is because non-equivalent norms have different Cauchy sequences and difference convergent sequences.

As an example, let $(X, \|\cdot \|_1)$ be a complete normed space and take $f:X \to \mathbb C$ to be an unbounded linear functional. Let $y \in X$ be such that $f(y) = 1$ (we can always find such $y$ by scaling $f$ if necessary). Next, define $T : X \to X$ by $T(x) = x - 2f(x)y$ for $x \in X$. Then $$T^2(x) = x-2f(x)y - 2f(x-2f(x)y)y = x - 2f(x)y - 2f(x)y + 4f(x)y =x$$ so $T^2 = I$. Thus, $$\| x \|_2 := \|T(x) \|_1, \,\,\,\, x \in X$$ defines a new norm which is not equivalent to $\| \cdot \|_1$ but which is still complete.

This construction is discussed here: https://www.researchgate.net/publication/226200984_Equivalent_complete_norms_and_positivity

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  • $\begingroup$ Thank you I will give the paper a look! $\endgroup$ – xk3 Dec 24 '16 at 22:22
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Let $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ be two equivalent norms on a vector space $X$. Then $\left( X, \lVert \cdot \rVert_1 \right)$ is complete if and only if $\left( X, \lVert \cdot \rVert_2 \right)$ is complete.

For a proof, we can find some positive real numbers $\lambda$ and $\mu$ such that $$\lambda \lVert x \rVert_1 \leq \lVert x \rVert_2 \leq \mu \lVert x \rVert_1 \ \mbox{ for all } \ x \in X.$$ Now suppose that $\left( X, \lVert \cdot \rVert_1 \right)$ is complete, and let $\left(x_n \right)_{n \in \mathbb{N}}$ be a Cauchy sequence in $\left( X, \lVert \cdot \rVert_2 \right)$. Then, given a real number $\varepsilon > 0$, we can find a natural number $N$ such that $$\lVert x_m - x_n \rVert_2 < \lambda \varepsilon$$ for all $m \in \mathbb{N}$ and $n \in \mathbb{N}$ such that $m > N$ and $n > N$. And so we have $$\lambda \lVert x_m - x_n \rVert_1 \leq \lVert x_m - x_n \rVert_2 < \lambda \varepsilon$$ for all $m \in \mathbb{N}$ and $n \in \mathbb{N}$ such that $m > N$ and $n > N$. Nos as $\lambda > 0$, so we can conclude that $$\lVert x_m - x_n \rVert_1 < \varepsilon$$ for all $m \in \mathbb{N}$ and $n \in \mathbb{N}$ such that $m > N$ and $n > N$, from which it follows that $\left( x_n \right)_{n \in \mathbb{N}}$ is a Cauchy sequence in $\left( X, \lVert \cdot \rVert_1 \right)$, and therefore this sequence converges in $\left( X, \lVert \cdot \rVert_1 \right)$ to some point $x \in X$. So we can find a natural number $K$ such that $$\lVert x_n - x \rVert_1 < \frac{\varepsilon}{\mu} \ \mbox{ and so } \ \lVert x_n - x \rVert_2 \leq \mu \lVert x_n - x \rVert_1 < \varepsilon$$ for all $n \in \mathbb{N}$ such that $n > K$, from which it follows that $\left( x_n \right)_{n \in \mathbb{N}}$ converges to $x$ in $\left( X, \lVert \cdot \rVert_2 \right)$, and the completeness of this norm space follows. And, similarly for the converse.

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  • $\begingroup$ This argument is fantastic thank you! May we conclude that if $\|\cdot\|_1$ and $\|\cdot\|_2$ are not equivalent on $X$ and $(X,\|\cdot\|_1)$ is complete, then $(X,\|\cdot\|_2)$ is not complete? $\endgroup$ – xk3 Dec 24 '16 at 20:42
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    $\begingroup$ @xk3 well, the result I've mentioned doesn't say this. Rather, we can say that if the two norms are equivalent, then the non-completeness w.r.t. one of these norms implies non-completeness w.r.t. the other. $\endgroup$ – Saaqib Mahmood Dec 24 '16 at 21:03

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