Proof that $\sum_i \lvert \alpha_i\rvert \leq |x|$, where $|x|$ is the norm

inequalitylinear algebranormed-spacesproof-explanation

On another post a user asked for a proof that if a map is linear then it is continuous. "Every linear mapping on a finite dimensional space is continuous"

I have pasted an image of the answer which I am having trouble understanding. In the last line he replaces $\sum_i \lvert \alpha_i\rvert$ with $|x|$, so it must either be equal or less for the conclusion to follow.

Question
Why is this sum of scalars equal or less than the norm?

Thanks

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Best Answer

Norms on finite-dimensional spaces are equivalent. You can check that $||x||_0 := \sum_i |\alpha_i|$ defines a norm on $X$ (in fixing that basis for $X$). Since the given norm $||\cdot||$ is another norm on $X$, a finite-dimensional space, it then follows that $||\cdot||_0$ and $||\cdot||$ are equivalent, i.e. there exists some constant $C>0$ such that $||\cdot||_0 \leq C ||\cdot||$. Adjusting your $\delta$ to be instead $\varepsilon / (CM)$ gets what you want

You can read something such as this post to learn more about this property of norm equivalence in finite-dimensional spaces

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