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Keeping the definition of norms in mind, given where a norm is defined (like the Euclidean space, or on a set of functions (like mappings from $C^{\infty} \to C^{\infty}$), or any other valid space) unique?

In other words, for instance, are the norms $||K|| = \sup_{I \subset R^n}(f - g) ; \ \ f,g:R^n \to R^m;\ \ f,g \in C^{\infty} $ or $||K||=\sqrt{x^2 + y^2}, \ \ x,y \in R \ $ unique where they are applied? Is there any other norm that passes the three properties of norms and yet different from ones I've mentioned?

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First of all, you need to make precise what it means for a norm to be unique; or rather when two norms are equal.

On an infinite dimensional vector space, there exist norms which are not equivalent (in the sense described here). Examples of such norms are the $p$-norms. For a function $f: \mathbb{R} \rightarrow \mathbb{R}$ s.t. the following integral exists, the $p$-norm with $1 \leq p < \infty$ is defined by $(\int \vert f(x) \vert^p dx )^{1/p}$. For any two different $p$ these norms are not equivalent.

However, on a finite dimensional (real) vector space, all norms are equivalent.

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    $\begingroup$ well if all possible defined norms are equal, doesn't it make that definition unique in a sense that no other different norm holds? $\endgroup$
    – cutus_low
    Oct 24 '19 at 20:52
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    $\begingroup$ It depends on what you mean by two norms being equal. Usually, one means that two norms are equal, if they induce the same topology. On a finite dimensional (real) vector space, this is the case. I.e. any two norms that are defined on $\mathbb{R}^n$ induce the same norm. For an infinite dimensional vector space (like the space of continuous real valued functions) this is not the case. $\endgroup$ Oct 24 '19 at 20:53

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