Is the l2-norm squared generated from an inner product?

Question

We know that the l2-norm is generated from an inner product by using the parallelogram law.

proof that l2 norm is generated from inner product

What about the squared of the $l_2$-norm? Is that generated from an inner product? I've tried to solve this using the parallelogram law but im not sure of my answer and i think it is no.

But in kernel ridge regression we use the $l_2$-norm squared, and according to the representer theorem the $l_2$-norm squared has to correspond to an inner product.

Will anyone be kind to show the proof to me? Many Thanks!

Answer 1

If $V$ is a real vector space, a norm on $V$ is a function $\|\cdot\| : V \to \mathbb R$, where $v \mapsto \|v\|$, such that the following properties holds:

• For any $v \in V$, $\|v\| \geq 0$; and $\|v\| = 0$ only if $v$ is the zero vector.
• For any $v \in V$ and $a \in \mathbb R$, $\|av\| = |a|\|v\|$.
• For every pair of vectors $v$ and $w$ in $V$, $\|v+w\| \leq \|v\|+\|w\|$.

Then, if $V = \mathbb R^n$ and we define $\|(x_1,\dots,x_n)\|$ by $x_1^2 + \cdots + x_n^2$, $\|\cdot\|$ is not a norm on $V$ since the second property doesn't holds for every $a \in \mathbb R$, unless $a = \pm 1$.