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We shall now learn how normed spaces give rise to new normed spaces. Suppose that $E, F$ are normed spaces. Then we can consider mappings $T: E \rightarrow F$ which are linear, i.e., which satisfy $$ T(f+g)=T f+T g \text { and } \quad T(\lambda f)=\lambda T f $$ for all $f, g \in E, \lambda \in \mathbb{K}$ ($\mathbb K$ could be $\mathbb R$ or $\mathbb C$). A linear mapping is also called a (linear) operator, and if the target space $F=\mathbb{K}$ is one-dimensional, they are called (linear) functionals. One can add linear mappings and multiply them by scalars by $$ (T+S) f:=T f+S f, \quad(\lambda T) f:=\lambda(T f) \quad(f \in E, \lambda \in \mathbb{K}) $$ and in this way the set of all linear mappings from $E$ to $F$ becomes a vector space. Is there a natural norm on that space?

If $E$ is infinite-dimensional, I think that then it is impossible to define a norm on the space of all linear mappings. Is it true? Why? How could I prove this?

Lely
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    If you consider continuous linear maps, then this is what you are looking for. – TheSilverDoe Jul 13 '23 at 10:19
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    Every vector space admits a norm, and spaces of linear maps are vector spaces. But unless you put some restrictions on what kind of linear maps you are considering there will not be any good norms (e.g. norms which make constructions like the map $L(E,F)\times E\to F, (T,v)\mapsto Tv$ continuous) – s.harp Jul 13 '23 at 11:06
  • see also https://math.stackexchange.com/questions/62778/does-every-mathbbr-mathbbc-vector-space-have-a-norm – daw Jul 13 '23 at 12:24

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Linear maps are homomorphisms of vector spaces. So if you have two vector spaces $E,F$, the set of all linear maps $E\to F$ is the set of homomorphisms between $E$ and $F$.

But you don't have just two vector spaces here. You have two normed vector spaces. Now your vector spaces carry a norm, whose purpose is to dote the spaces with a metric space structure where the vector-space operations are continuous. If you now consider the set of all linear maps $E\to F$, which of course you can do, you are allowing maps which are not compatible with said structure (namely, discontinuous linear maps). As mentioned in the comments, you can put a norm in any vector space over $\mathbb R$ or $\mathbb C$ simply by choosing a basis and defining the norm of an element by taking the sum of the absolute values of the coefficients. But this norm would have zero relation with the topological structure of $E$ and $F$, so it is hard to imagine what purpose it would serve.

Instead, one usually consider the set of continuous linear maps between $E$ and $F$. These would be precisely the homomorphisms between $E$ and $F$, considered as topological vector spaces. By an exercise that is present in every Functional Analysis class/text, these are precisely the bounded linear maps. That is, a linear map $T:E\to F$ is continuous if and only if $T$ is bounded, which means that there exists $c>0$ such that $$ \|Tx\|\leq c\|x\|,\qquad x\in E. $$ This property suggests how to define a canonical norm on $B(E,F)$, the vector space of bounded linear maps $E\to F$. This is $$ \|T\|=\inf\{c:\ \|Tx\|\leq c\|x\|,\ x\in E\}. $$ It is not hard to show that $$ \|T\|=\sup\{\|Tx\|:\ x\in E,\ \|x\|=1\}=\sup\Big\{\frac{\|Tx\|}{\|x\|}:\ x\in E,\ x\ne0\Big\}. $$ This norm on $B(E,F)$ is called the operator norm.

Martin Argerami
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  • This means that if we consider a sequence of linearly independent vectors in $E$, we can find linear maps that map this sequence to arbitrarily large values? (This could imply that we cannot find a uniform upper bound for the norms of linear maps, making it impossible to define a consistent norm on the space of all linear maps, right?) – Lely Jul 15 '23 at 16:11
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    Yes. $ \ \ \ \ \ \ \ $ – Martin Argerami Jul 15 '23 at 17:04