# Showing any linear operator $T : X \to Y$ is bounded, where $X$ is a finite dimensional normed vector space, and $Y$ any normed vector space.

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?

• Hint: What theorem do you know about norms on a finite-dimensional vector space? Dec 16, 2018 at 11:02
• Norms on a finite-dimensional vector space are equivalent? Dec 16, 2018 at 11:14
• Yes! So what can you infer, if you know that the original norn and the one you defined are equivalent? Dec 16, 2018 at 11:16
• $∥Tx∥ \leq ∥x∥ + ∥Tx∥ = ∥x∥_0$ Dec 16, 2018 at 11:25
• 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. Dec 16, 2018 at 11:41

$$||T∥_2 = \sup \limits_{x \neq 0} \frac{∥Tx∥_1}{∥x∥_1} = \sup \limits_{x \neq0} \frac{∥( x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_1} = \sup \limits_{x \neq0} \frac{|x_1|+…+|x_n|}{|x_1|+…+|x_n|}= 1$$
• Why are you using the $1$ norm for $x$ if $X$ is equipped with the Euclidean norm? Dec 16, 2018 at 16:37
• Because $Y := \ell_1(\mathbb{N})$ Dec 16, 2018 at 16:39
• But for $x$ you have to look at the norm of $X$. You should use the norm of $Y$ just for $Tx$ Dec 16, 2018 at 16:42
• 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 Dec 16, 2018 at 16:50