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?

  • 1
    $\begingroup$ Why do you think that $\ell_1$ is a closed subspace of $\ell_2?$ $\endgroup$
    – mfl
    Commented Aug 2, 2018 at 8:56

1 Answer 1


$\ell_1$ 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.


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