0
$\begingroup$

I'm having some trouble understanding the proof to the following proposition:

Let $E,F$ be two normed spaces, such that $E$ has finite dimension. If $f:E\to F$ is linear, then it is continuous.

The proof is the following:


Let $e_1,...,e_n$ be a basis for $E$. Then, for any $x\in E$, we have that: $x=\sum_{i=1}^n x_ie_i$. Then $$f(x)=\sum_{i=1}^n x_if(e_i)$$ And so we get the following inequality: $$||f(x)||\leq b||x||_\infty$$ with $b=\sum_{i=1}^n||f(e_i)||$.

Because in $E$ all norms are equivalent, we have that $f$ is continuous.


I don't get why the fact that all norms are equivalent allows us to conclude that $f$ is continuous. Why is it so?

$\endgroup$
3
  • 1
    $\begingroup$ $\|f(x)\|_F\leq b\|x\|_{\infty}\leq B\|x\|_E$ for some possibly larger constant $B$, where the last step uses equivalence of norms. This inequality being true for all $x\in E$ is equivalent to continuity of $f$. $\endgroup$
    – peek-a-boo
    Dec 22, 2022 at 19:05
  • $\begingroup$ I don't get how it's equivalent to the continuity of $f$ @peek-a-boo $\endgroup$ Dec 22, 2022 at 19:07
  • 2
    $\begingroup$ this you need to work out yourself, or look in a standard textbook, or look on this site; it has definitely been asked several times. By linearity, it tells you $\|f(x)-f(y)\|\leq B\|x-y\|$, which gives you the Lipschitz condition, and hence continuity (the converse is only slightly trickier, which you should work out). $\endgroup$
    – peek-a-boo
    Dec 22, 2022 at 19:08

1 Answer 1

3
$\begingroup$

We know that a $f$ is continuous on $E$ if

$\forall y_0 \in E \ \forall \epsilon >0 \ \exists \delta >0 : \forall y$ such as $||y-y_0||_E<\delta$ implies $ ||f(y)-f(y_0)||_F<\epsilon $

Since $f$ is linear we can rewrite $||f(y)-f(y_0)||_F=||f(y-y_0)||_F$ and consider the following definition of continuity in this case : (by taking $x=y-y_0$)

$\forall x \in E \ \forall \epsilon >0 \ \exists \delta >0$ such as $||x||_E<\delta$ implies $||f(x)||_F<\epsilon $

We have the inequality that you mentionned and, $||f(x)||_F\leq b||x||_\infty\leq B||x||_E$ because of the equivalence of the norms.

Then, if we consider an $\epsilon$ you just have to pick $\delta\leq \frac{\epsilon}{B} $ then $||f(x)||_F\leq B||x||_E \leq \delta B \leq \frac{\epsilon B}{B} \leq \epsilon$

Thus, $f$ is continuous.

$\endgroup$
2
  • $\begingroup$ your answer is perfect. just one question: When we set $\delta \leq \epsilon/B$, how do you then conclude that $B||x||_E \leq \delta B$? $\endgroup$ Dec 23, 2022 at 13:13
  • $\begingroup$ We set the condition $||x||_E\leq\delta$ in the definition, then $B||x||_E\leq\delta B$ with $B>0$ $\endgroup$
    – Camolistia
    Dec 24, 2022 at 8:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .