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I had to prove the following statement:

Let $E$, $F$ be normed spaces. If $E$ is finite dimensional then $\mathcal{L}(E,F)=\mathcal{L}^1(E,F)$.

($\mathcal{L}^1(E,F)$ represents the space of bounded linear maps $E\rightarrow F$)

I kind of struggled and when I checked the proof in my textbook it had nothing to do with the one I wrote (and I still don't understand the one in the book...).

Is mine correct?

proof:

We proceed by induction.

Consider the proposition, $\forall n\in\mathbb{N}^*:\bigg(\dim E = n \Rightarrow \mathcal{L}(E,F)=\mathcal{L}^1(E,F)\bigg)$

If $E$ is a one dimensional normed space then $E=\langle e \rangle$ for some $e\in E$.

Let $f\in\mathcal{L}(E,F)$

Recall that we call $f$ bounded if $\|f\|=\sup_{\mathbb{R} \cup \lbrace \infty \rbrace}\bigg\lbrace \frac{\| f(v) \|}{\|v\|} \bigg | v\in E \wedge v\neq0 \bigg\rbrace < \infty$

So for all $v\in E$, $v=\lambda e$ for some $\lambda \in \mathbb{C}$.

So $\frac{\| f(v) \|}{\|v\|}=\frac{\|f(e)\|}{\|e\|}$ is a constant hence $f$ is bounded.

Suppose $\exists n\in\mathbb{N}^*:P_n$ is true.

Let $V$ a $n+1$ dimensional space.

Then we can write $V$ as $V=E\oplus \langle e \rangle$ where $\dim E = n$ and $e\in V$.

Let $f\in\mathcal{L}(E,F)$.

Then $f|_E$ and $f|_{\langle e \rangle}$ are bounded so immediatly we have that $f$ is bounded.

So $P_{n+1}$ is true.

Therefore the proposition is true of all $n\in\mathbb{N}^*$.

Anne Bauval
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Julia
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  • @AnneBauval it was a mistake I meant domain :( – Julia Dec 30 '22 at 18:03
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    Your proof would be ok if its heart ("so immediatly we have that $f$ is bounded") was carefully detailed. For an easier one, see https://math.stackexchange.com/a/113008/386889 – Anne Bauval Dec 30 '22 at 18:06
  • Your proof is incorrect. Let a normed space $X$ be represented as $X=Y\oplus \mathbb{C}x_0$ for some $x_0\notin Y.$ Let an operator $T$ acts by $T(y+ax_0)=ax_0.$ Then $T$ is bounded on $Y$ and on $\mathbb{C}x_0$ but it may happen that $T$ is unbounded on $X.$ You should make use of the fact that $Y$ is finite dimensional but there is no such argument in your reasoning. – Ryszard Szwarc Dec 30 '22 at 19:26

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