Given a Banach space X and the fact that all norms are equivalent, does it follow that dimension of X is finite? The converse statement is proved in many books however I didn't find this statement neither proved or disproved anywhere. Does there exist any relevant book where this is described? Thank you for any links or answers.
Filip Soudský
Dear Filip Soudský
On every infinite-dimensional linear space (real or complex) there exist non-equivalent norms. Usually this statement is given as an exercise, like for example Exercise 4 in Section 10.2.1 of my book ``A Course in Functional Analysis and Measure Theory'' https://www.springer.com/us/book/9783319920030#aboutBook
Let me give here a possible solution. Let $X$ be an infinite-dimensional linear space. Select a Hamel basis $e_t, t \in T$ for $X$ (see the definition and the existence theorem in sections 5.1.1 and 5.1.3 of the same book). Every element $x \in X$ is uniquely representable as a linear combination of $e_t, t \in T$, that is in the form
$$\sum_{t \in T} a_t(x) e_t,$$
where all $a_t(x)$ are scalars and only finite number of $a_t(x)$ are different from zero.
Define two norms: $||x|| = \sum_{t \in T} |a_t(x)|$ and
$|||x||| = \max_{t \in T} |a_t(x)|$. Both norms are correctly defined, because in the corresponding sum and max only finitely many nonzero coordinates are involved,
but these two norms are not equivalent.
Best, Vladimir Kadets
I recommend you to read the fifth chapter of the following book: A Course in Functional Analysis and Measure Theory
Authors: Kadets, Vladimir,
The answer is yes: if all norms are equivalent on a linear space $E$, then $E$ is finite dimensional. In fact on every infinite dimensional space, one can consider infinitely many norms (just take norms as by Professor Kadets, but with different weights).
One can also consider a norm $|| ||$ on $E$ and construct a discontinuous functional $f$ on $E$, then $|||x||| := ||x|| + |f(x)| is also a norm which is not equivalent to the given one.
All the best
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