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Let $E,F$ be normed vector spaces.

Consider the mapping $\sum_{k=1}^n y_i \otimes f_i: x \mapsto \sum_{k=1}^nf_i(x)y_i $, where $x \in E, y_i \in F, f_i \in E'$. Show that the mapping above is a continous linear operator with finite dimensional image.

My approach: The linearity follows obviously, since we have a sum of (continous) linear functionals. By the same argument, I get the continuity.

Definition: An operator $T$ is of finite rank (or finite dimensional image) if its range has finite domension.

This means I have to look at $T(E)$, and show that for $w \in T(E)$, w can be written using a finite amount of basis vectors. So let $B$ be a basis of $T(E)$, and let $w \in T(E)$. Then we can write $w$ as, $w=\sum_{k=1}^n f_i(x)y_i$.

I don't know how to continue the argument.

Question/request: Regarding this topic, there are some things that I think I don't really understand (or am confused about in general), but I am not sure what does are. So if possible, a detailed solution would be very helpful.

wanymose
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Hint: you don’t need a finite basis directly — it suffices to find a finite spanning set (any spanning set can be reduced to a basis).

Second hint: you have a very conveniently laid out spanning set for you.

Solution: note that the set of $y_i$ form a finite spanning set for the image, and so the image is finite dimensional.

Nicholas Priebe
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  • But for $x \in E$ I have that $T(x)=\sum_{k=1}^n f_i(x)y_i$, as far as I did understand it those $y_1,...,y_n$ are just some elements in $F$. So if I take some other element of $E$, lets call it $\tilde{x}$, I would get $T(\tilde{x})=f_i(x)\tilde{y}_i$ for some other. $\tilde{y}_i \in F$, I don't see how this would get me a finite spanning set. – wanymose Jun 26 '23 at 01:09
  • Wait — you’re not fixing the $y_i$ at the start? your notation suggests to me that the $y_i$ are fixed vectors built into the function. If the $y_i$ vary by input, how are they selected? – Nicholas Priebe Jun 26 '23 at 01:16
  • I don't know, that's just an exercise. There is no more text in it. If the $y_i$'s are fixed, then it is obvious that ${y_1,...,y_n}$ is a spanning set. – wanymose Jun 26 '23 at 01:22
  • Thank you! It seems like I just misunderstood the exercise – wanymose Jun 26 '23 at 01:23