# Proving that $f:E\to F$ is continuous if it's linear

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?

• $\|f(x)\|_F\leq b\|x\|_{\infty}\leq B\|x\|_E$ 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 $f$. Dec 22, 2022 at 19:05
• I don't get how it's equivalent to the continuity of $f$ @peek-a-boo Dec 22, 2022 at 19:07
• 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). Dec 22, 2022 at 19:08

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.

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