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I have a question and I am fully aware it is "trivial". I wonder the following: If we minimize $\|x\|^{2}$ at the same time also $\|x\|$ becomes minimal. This is trivial insofar as it is very basic intuition. What I have realized though is that I would not have any idea how to proof that formally. How is that done?

With best regards, Miau

MiauPiau
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We need some context here, what is $||.||$? I am assuming it means Euclidean norm.

If $a,b\geq 0$ are real numbers such that $a^2 \leq b^2$ then $a \leq b$. Now apply this to the positive numbers $||x||^2$ and $||y||^2$ for some vectors $x,y$ where $x$ minimizes $||.||^2$.

fretty
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  • Yes, eulidean norm was meant and the context was related to Support Vector Machines where it is stated that maximizing $\frac{1}{|x|}$ would be equivalent to minimizing $\frac{1}{2}|x|^{2} $. Thanks – MiauPiau Feb 06 '13 at 20:35
  • In fact, the converse is true as well; both are due to the monotonic behavior. – Squirtle Jul 19 '13 at 04:38
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This is due to the fact that $f(x) = x^2$ is an increasing function on $[0, \infty)$. If $f$ is restricted to some domain $A \subseteq [0, \infty)$, then the minimum of $x$ corresponds to the minimum of $f(x) = x^2$.

ferson2020
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  • of course..so obvious, if we had an $x_{1}$ for which $f(x_{1})$ = minimum of A, but $\exists x_{2}$ with $x_{2} < x_{1}$ would imply $f(x_{2}) > f(x_{1})$ which would be in contradiction to the assumption that f is an increasing function. thanks! – MiauPiau Feb 06 '13 at 20:32