Linear function between normed spaces is continous.

Question

Let $A: X \rightarrow Y$ be linear function between normed spaces $X$, $Y,\ dimX \lt \infty$. Prove that A is continous.

Since $A$ is linear I know that there exists $k \in \mathbb{R}$ such that $||A(x)||\le k||x||$ for all $x \in X$.

So now I check $||A(x)-A(y)||=||A(x-y)||\le k||x-y||=k\delta=\epsilon$ supposing $||x-y||\lt \delta=\epsilon/k$

Is it a correct proof?

The same question was posted here: https://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous — Olvi, Apr 20 '17 at 10:45
The proof isn't correct since you're assuming your own thesis. In fact, usually the definition of continuity is given by your second line — Exodd, Apr 20 '17 at 10:46
It looks ok if you properly explain the second line. For example, the norm is continuous, the unit ball in a finite dimensional space is compact plus Weierstrass. — A.Γ., Apr 20 '17 at 10:52

score 1 · Accepted Answer · answered Apr 20 '17 at 10:44

Since $X$ is finite dimensional space, all norms are equivalent in X, so let $(e_i)_i$ be a finite algebriac basis of $X$ and we equipe $X$ by the norme: $$ \|x:=x_1e_1+\dots x_ne_n\|=\sum_i |x_i| $$ so $$ \|Ax\|=\|\sum_i x_i Ae_i\|\leq\sum_i \|x_i Ae_i\| \leq\max_i \|Ae_i\|\sum_i |x_i|=\alpha \|x\| $$ Where $\alpha=\max_i \|Ae_i\|<\infty$ So $A$ is continuous

Linear function between normed spaces is continous.

1 Answers1