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Let $X$ be a finite dimensional normed vector space and $Y$ an arbitrary normed vector space. Show that any linear operator $T : X \to Y$ is bounded.

I got the hint to first show that $\| x\|_0 := \| x \| + \| Tx\|$, $x \in X$, defines a norm on $X$, but I do not know how this should help me.

Further I should calculate $\|T\|$ for where $X = K^n$, equipped with the Euclidean norm $\|\cdot\|_2$, $Y := \ell_1(\mathbb{N})$ and $Tx := (x_1,\ldots,x_n,0,0,\ldots) \in \ell_1(\mathbb{N})$, for all $x = (x_1,\ldots,x_n) \in K^n$.

Can please someone help?

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  • $\begingroup$ Hint: What theorem do you know about norms on a finite-dimensional vector space? $\endgroup$
    – Aphelli
    Dec 16, 2018 at 11:02
  • $\begingroup$ Norms on a finite-dimensional vector space are equivalent? $\endgroup$ Dec 16, 2018 at 11:14
  • $\begingroup$ Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent? $\endgroup$
    – Aphelli
    Dec 16, 2018 at 11:16
  • $\begingroup$ $∥Tx∥ \leq ∥x∥ + ∥Tx∥ = ∥x∥_0$ $\endgroup$ Dec 16, 2018 at 11:25
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    $\begingroup$ I am sorry, I must have misunderstood your first comment and confused you further. You do have $\|Tx\| \leq \|x\|_0$, thus the operator is bounded. $\endgroup$
    – Aphelli
    Dec 16, 2018 at 11:41

1 Answer 1

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Regarding the operator norm: I was thinking

$ ||T∥_2 = \sup \limits_{x \neq 0} \frac{∥Tx∥_1}{∥x∥_1} = \sup \limits_{x \neq0} \frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_1} = \sup \limits_{x \neq0} \frac{|x_1|+…+|x_n|}{|x_1|+…+|x_n|}= 1 $

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    $\begingroup$ Seems correct ! $\endgroup$
    – Ramanujan
    Dec 16, 2018 at 12:18
  • $\begingroup$ Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm? $\endgroup$
    – user289143
    Dec 16, 2018 at 16:37
  • $\begingroup$ Because $Y := \ell_1(\mathbb{N})$ $\endgroup$ Dec 16, 2018 at 16:39
  • $\begingroup$ But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$ $\endgroup$
    – user289143
    Dec 16, 2018 at 16:42
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    $\begingroup$ True. So $ ||T∥_2 = \sup \limits_{x \neq 0} \frac{∥Tx∥_1}{∥x∥_2} = \sup \limits_{x \neq0} \frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = \sup \limits_{x \neq0} \frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{\frac{1}{2}}}= ? $ and I thought I had figured it out $\endgroup$ Dec 16, 2018 at 16:50

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