Square rooting results

Question

This is probably a very silly question, but say I have $∥f∥_1^2$ (I'm trying to prove an inequality) and I square root it, do I get $-{∥f∥_1}$ and $∥f∥_1$ or just $∥f∥_1$ due to the definition of the 1 norm (beign non-negative)?

Thank you

Answer 1

This doesn't have much to do with the norm.

The square root symbol $\sqrt{x}$, by convention, represents a function whose domain is $[0,\infty)$ and whose range is $[0,\infty)$. So the input $x$ must satisfy $x \ge 0$, and the output satisfies $\sqrt{x} \ge 0$.

So if, as you say, you want to compute the square root of $\|f\|_1^2$, the answer will be $\|f\|_1$.

Answer 2

$\sqrt{a^2} = |a|$

If $a \ge 0$ then $\sqrt{a^2} = |a| =a$.

But if $a \le 0$ then $\sqrt{a^2} = |a| = -a$.

You have to be careful.

If you want to solve $(x+2)^2 = 9$ it is true that that means

$\sqrt{(x+2)^2} = \sqrt{9}$ which means

$|x+2| =3$ (but notice its the ABSOLUTE VALUE of $x+2$ that is equal to $3$ and not $x+2$ itself.

SO that means EITHER $|x+2| = x + 2 =3$ and $x = 1$ OR it could mean that $|x+2|=-(x+2)=3$ or in other words that $-x-2=3$ or we could say $x+2=-3$ and therefore $x=-5$.

So $(x+2)^2 = 9$ has two possible solution $x =1$ or $x=-5$.

THis $x^2 = M$ and $M = k^2$ for some positive $k$; so $\sqrt {x^2} = \sqrt M$ and $|x| = k$ so $\pm x = k$ so $x= \pm k$ is so common we usually just state:

$(x+2)^2 = 9$

$(x+2) = \pm \sqrt 9 = \pm 3$.

......

But by definition $∥f∥_1$ is always positive (or $0$).

So if you have $∥f∥_1^2 = 25$ then

$\sqrt{∥f∥_1^2} = \sqrt {25} = 5$ so

$|∥f∥_1 | = 5$.

But because $∥f∥_1$ is never negative we know

$|∥f∥_1| = ∥f∥_1$.

So we have $\sqrt{∥f∥_1^2} = ∥f∥_1$ .