# What is wrong with the proof of $\ell_1$ is nonreflexive?

Recall that $$\ell_1 = \{(x_n)_{n\in\mathbb{N}}: \sum_{n=1}^\infty |x_n|<\infty\}$$ and $$\ell_2 = \{(x_n)_{n\in\mathbb{N}}: \sum_{n=1}^\infty |x_n|^2<\infty\}.$$

Known facts:

$$(1)$$ $$\ell_1$$ and $$\ell_2$$ are Banach spaces. In particular, $$\ell_1$$ is closed.

$$(2)$$ $$\ell_2$$ is reflexive.

$$(3)$$ $$\ell_1\subseteq \ell_2$$.

$$(4)$$ Closed subspace of a reflexive space is again reflexive.

Therefore, $$\ell_1$$ is reflexive.

But it is a well-known fact that $$\ell_1$$ is never reflexive. What is wrong with the above arguments?

• Why do you think that $\ell_1$ is a closed subspace of $\ell_2?$
– mfl
Commented Aug 2, 2018 at 8:56

$\ell_1$ is not closed as a subspace of $\ell_2$: when you say $\ell_1$ is closed, you mean that it is closed with respect to the $\lVert \cdot \rVert_1$ norm, while when we consider it as a subspace of $\ell_2$ we are talking about the $\lVert \cdot \rVert_2$ norm.
To see that $\ell_1$ is not closed in $\ell_2$, just take an element $x$ of $\ell_2$ which is not in $\ell_1$, as for example is $(\frac{1}{n})_n$, and build a sequence of elements in $\ell_1$ which converge to $x$ in $\lVert \cdot \rVert_2$ norm. It should be easy for you do find many such examples.