How to check triangle inequality in definition of norm

linear algebranormed-spacestriangle-inequalityvector-spaces

I’m trying to solve a problem where I am supposed to check if the formula defined below is qualified to be a norm(that is, check if definition of norm works fine with the formula).

$||(u,v)||=\sqrt{||u||^2_U+||v||^2_V}$ , where U, V are vector space over $R$ and $||\cdot||_U$, $||\cdot||_V$ are norms defined on U, V respectively.

I know that(at least according to my book) for a norm to be defined on a vector space(here, just assumed the vector space to be some arbitrary V),

  1. $||u||\geq0$, for all u, and $||u||=0$ iff $u=0$
  2. $||au||=|a|\times||u||$, for all $a\in R$
  3. $||u+v|| \leq||u||+||v||$, (triangle inequality).

must hold.

Fortunately, I proved for 1 and 2. But the problem is I am not sure how to check 3 also work for the formula defined above. I tried to prove it by solving for $||(u,v)+(x,y)||$ but all the squares and roots in the operation got me confused and I was unable to derive 3 from the formula. I am guessing somehow I have to cleverly used the triangle inequality defined in $||\cdot||_U$, $||\cdot||_V$, but I am not sure how…

Could anyone help me with this? Thank you.

Best Answer

Let us start with a useful inequality: for every $a,b,c,d > 0$ we have $$ ac+bd \leq \sqrt{a^2+b^2}\sqrt{c^2+d^2}. $$ Indeed, by squaring both sides we get $$ a^2c^2 + 2abcd + b^2d^2 \leq (a^2+b^2)(c^2+d^2) \iff 2abcd \leq b^2c^2+a^2d^2 \iff 0 \leq (bc-ad)^2. $$

Now, using the definition and the above with $a = \lVert u \rVert_U$, $b = \lVert v \rVert_V$, $c = \lVert u' \rVert_U$, $d = \lVert v' \rVert_V$ we get \begin{align*} \lVert (u,v) + (u',v') \rVert^2 &= \lVert u + u' \rVert_U^2 + \lVert v + v' \rVert_V^2 \leq (\lVert u \rVert_U + \lVert u' \rVert_U)^2 + (\lVert v \rVert_V + \lVert v' \rVert_V)^2 = \\ &= \lVert u \rVert_U^2 + \lVert u' \rVert_U^2 + 2 \lVert u \rVert_U \lVert u' \rVert_U + \lVert v \rVert_V^2 + \lVert v' \rVert_V^2 + 2 \lVert v \rVert_V \lVert v' \rVert_V \\ &\leq \lVert u \rVert_U^2 + \lVert u' \rVert_U^2 + \lVert v \rVert_V^2 + \lVert v' \rVert_V^2 + 2 \sqrt{\lVert u \rVert_U^2 + \lVert v \rVert_V^2}\sqrt{\lVert u' \rVert_U^2 + \lVert v' \rVert_V^2} \\ &= (\sqrt{\lVert u \rVert_U^2 + \lVert v \rVert_V^2} + \sqrt{\lVert u' \rVert_U^2 + \lVert v' \rVert_V^2})^2. \end{align*}

Taking the square root on both sides leads to the triangular inequality for the norm $\lVert \cdot \rVert$

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