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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

thomas_stafford_22
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2 Answers2

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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 $\|f\|_1$.

Lee Mosher
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  • So does that mean I would only get $∥f∥_1$? – thomas_stafford_22 Jan 05 '20 at 21:54
  • Sorry to ask such a simple question, but with you're saying, say we take root(25), we can have -5 which is not more than or equal to 0 which you said it has to be – thomas_stafford_22 Jan 05 '20 at 21:56
  • What I said is regarding the square root symbol, so $\sqrt{25}=5$. If you were asking about the roots to the equation $x^2=25$, then the answer would be different, those two roots are $\sqrt{25}=5$ and $-\sqrt{25}=-5$. The square root symbol itself represents a well-defined function. Writing $\sqrt{25}=$"$+5$ and $-5$" is not correct. – Lee Mosher Jan 05 '20 at 21:58
  • Wow I can't believe I've never noticied that before - thank you – thomas_stafford_22 Jan 05 '20 at 22:00
  • see: https://math.stackexchange.com/questions/3310604/can-square-roots-be-negative/3310697#3310697 – Adam Rubinson Jan 05 '20 at 22:28
  • A subtle (be important and REAL) thing to note is that $x^2=25$ has two solutions. But $x =\sqrt{25}$ has only one. $\sqrt{}$ is defined to always be non-negative. So $\sqrt{5^2} = 5$. And $\sqrt{(-5)^2} = 5$. So $\sqrt{a^2}\ne a$ if $a$ is negative. Instead $\sqrt{a^2} = |a|$. ... Important to note that if $a < 0$ then $|a| =-a$. And $-a > 0$. This is NOT a contradiction or weird. The negative sign does NOT mean the value is negative. $-(-5) =5$ and $-(negative)$ is positive. So a negative sign, $-a$ doesn't mean negative. It means opposite of what $a$ is. – fleablood Jan 06 '20 at 01:07
  • the reason $\sqrt{∥f∥_1^2} = ∥f∥_1$ is because $∥f∥_1$ is never negative and $\sqrt{}$ is never negative. – fleablood Jan 06 '20 at 01:08
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$\sqrt{a^2} = |a|$

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 $\sqrt{∥f∥_1^2} = ∥f∥_1$ .

fleablood
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