Continuity of an $\mathbb {R}^2$ function Let $f$ be an $\mathbb{R}^2$ endomorphism and $N:\mathbb{R}^2\to\mathbb{R}^+$ 
defined by $$\forall u \in \mathbb {R }^2, N(u) = ||f(u)|| $$
I need to show $N$ is continuous.
The problem is that $N$ is only a seminorm, otherwise it would have verified Lipschitz criterion and I'd be done.
Thanks for your help.
 A: Let's prove that a linear function is continuous. Recalling the definition:

A function $f:\Bbb{R}^n\to\Bbb{R}^k$ is continuous iff for each $x\in\Bbb{R}^n$ and for each $\epsilon>0$ there exists a $\delta>0$ such that for all $y\in\Bbb{R}^n$ with $||y-x||<\delta$ we have $||f(y)-f(x)||<\epsilon$.

Now if $f$ is linear, then it takes the form $f(x)=Ax$ for some $k\times n$ matrix $A$. Then
$$||f(y)-f(x)||=||Ay-Ax||=||A(y-x)||,$$
and we want to show that this is less than $C||y-x||$ for some constant $C$ dependent on $A$ only. This would finish the proof, because for any $\epsilon$, if we let $\delta=\epsilon/C$ we have that
$$||f(y)-f(x)||=||A(y-x)||\le C||y-x||<C\delta=\epsilon,$$
and so $f$ is continuous.
But let's back up the claim that for all $x\in\Bbb{R}$, $||Ax||\le C||x||$ for some $C$. (The smallest upper bound for $C$ here is conventionally called the induced matrix norm $||A||$.) An easy-to-prove upper bound here is the Frobenius norm, defined by $||A||_F^2=\sum_{i=1}^k\sum_{j=1}^nA_{ij}^2$.
Writing out all these norms as sums (and squaring both sides for convenience):
$$||Ax||^2=\sum_{i=1}^k\left(\sum_{j=1}^nA_{ij}x_j\right)^2$$
$$||A||_F^2||x||^2=\left(\sum_{i=1}^k\sum_{j=1}^nA_{ij}^2\right)\sum_{k=1}^nx_k^2=\sum_{i=1}^k\left(\sum_{j=1}^nA_{ij}^2\cdot\sum_{k=1}^nx_k^2\right)$$
We wish to prove that $||Ax||^2\le ||A||_F^2||x||^2$, but given the above expressions, this reduces to
$$\left(\sum_{j=1}^nA_{ij}x_j\right)^2\le\sum_{j=1}^nA_{ij}^2\cdot\sum_{k=1}^nx_k^2,$$
which is exactly the famous Cauchy–Schwarz inequality. This completes the proof.

Now that we know that linear functions are continuous, let's return to the problem. We want to know that an endomorphism of $\Bbb{R}^2$, i.e. a linear function $f:\Bbb{R}^2\to\Bbb{R}^2$, composed with the norm operation $||\cdot||:\Bbb{R}^2\to\Bbb{R}$, is continuous. If it is accepted that the norm operation is continuous and the composition of continuous functions is continuous, then the above proof is sufficient to prove that $N(x)=||f(x)||$ is continuous.
Although the above was proven in full generality of $\Bbb{R}^n\to\Bbb{R}^k$, for application to the problem we need only $\Bbb{R}^2\to\Bbb{R}^2$, and the only linear functions $\Bbb{R}^2\to\Bbb{R}^2$ are functions of the form $f(x)=Ax$ for $A$ a $2\times2$ matrix; this yields the representation $f(x)=(ax_1+bx_2,cx_1+dx_2)$. Note that these functions (choosing arbitrary constant $a,b,c,d\in\Bbb{R}$) are the only endomorphisms of $\Bbb{R}^2$.
