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I am working with Banach Spaces, which are complete Normed Vector Spaces (NVS).

The norm on a NVS $(E, ||\cdot||_1)$ defines a metric which in turn defines a topology.

Now let us consider $F \subset E$, a set that is itself a NVS with a different norm $||\cdot||_2$.

Can this space $F$ be considered a subspace of $E$? If we suppose that the norm $||\cdot||_2$ cannot be defined in the whole $E$, Does the definition of subspace of a NVS require the norm to be the same?

I need to clarify this concept, because I want to be rigorous. I am facing this situation, in which I have $E \subset F$, $E$ is not closed in $F$ in the topology induced by $||\cdot||_1$, but both are Banach Spaces (i.e both are complete), obviously considering them with their respective norms.

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    $\begingroup$ Same norm is required. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 14 '16 at 11:36
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    $\begingroup$ in finite dimensional vector spaces all the norms are equivalent : if a sequence converges for some norm it will converge for all the other norms. this is not true anymore in infinite dimensional vector spaces, that's why we say that a Banach space $(E,\|.\|_1)$ is completely different from an other $(E,\|.\|_2)$ on the same vector space. and that's the problem : when you study the Fourier transform you jump from $L^1(\mathbb{R})$ to $L^\infty(\mathbb{R})$ and $L^2(\mathbb{R})$ or even $L^2([a,b])$ and their respective duals, and that's complicated. $\endgroup$ – reuns Apr 14 '16 at 11:44

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