# Let $T: V \to W$ be a linear map. Let $V$ be finite dimensional. Then $T$ is continuous.

Prove the following, without using that any two norms on a finite dimensional vector space are equivalent.

Let $T: V \to W$ be a linear map between normed spaces. Let $V$ be finite dimensional. Then $T$ is continuous.

My attempt:

I can prove that $T$ is continuous iff $T$ is continuous in $0$.

So, it suffices to show:

$$\forall \epsilon > 0: \exists\delta > 0: \forall v \in V: \Vert v \Vert <\delta \implies \Vert Tv\Vert < \epsilon$$

Let $E := \{e_i\}_i^n$ be a basis of $V$. Write $v = \sum v_i e_i$

Let $\epsilon > 0$. Choose $\delta := ???$

If $v \in V$ satisfies $\Vert v\Vert < \delta$, then $\Vert Tv\Vert = \Vert \sum v_iT(e_i) \Vert \leq \sum |v_i| \Vert Te_i\Vert< M\sum |v_i|$

where $M := \max_i^n\Vert Te_i \Vert$

However, I don't know how to handle the $\sum |v_i|$. How can I make this small, given that $\Vert v \Vert < \delta?$

• I am not quite sure myself but cant you just use the fact that $||v||<\delta$ and then put the sum at the end $\leq M\sum \delta$ then introduce a $\delta_1$ such that the whole thing will be less than $\epsilon$? EDIT: As the person showed below. Simple use the definition of a norm. – ʎpoqou Feb 18 '18 at 16:22
• The answer below assumes there is an orthogonal basis, which uses Gramm-schmidt, and I do not have acces to an inner product, unless you can equip any finite dimensional space with an inner product? – user370967 Feb 18 '18 at 16:29
• @Math_QED You can definitely endow any finite dimensional real vector space with an inner product, let $(e_1,\ldots,e_n)$ be a basis and let define: $$\left\langle\sum_{i=1}^nx_ie_i,\sum_{j=1}^ny_je_j\right\rangle=\sum_{k=1}^nx_ky_k,$$ then $\langle\cdot,\cdot\rangle$ is an inner product and $(e_1,\ldots,e_n)$ is orthonormal. – C. Falcon Feb 18 '18 at 16:36
• So that solves the problem in the answer below? – user370967 Feb 18 '18 at 16:36
• Not really, because your spaces are already endowed with given norms and you are not allowed to use the equivalence of norms... – C. Falcon Feb 18 '18 at 16:38

Suppose $T: V \rightarrow W$ is not continuous and $V$ is finite dimensional. Then there is a sequence $(x_n)_n$ with $||x_n||=1$ and $||Tx_n||\geq n$ for all $n\in\mathbb{N}$. Let be $(e_i)_{i=1}^N$ a basis of $V$. Bolzano-Weierstraß implies the existence of a subsequence $(x_{n_k})_k=(\sum\limits_{i=1}^N a_{i,n_k} e_i)$ converging towards $x=\sum\limits_{i=1}^N a_i e_i \in B_1(0):=\lbrace y \in V \; | \;||y||\leq 1\rbrace$ (which is compact in $V$ iff $V$ is finite dimensional) and with $$|a_{i,n_k}-a_i| \rightarrow 0$$ for all $i=1,...,N$. But we also have $$0\leq ||Tx||\leq \sum\limits_{k=1}^N |a_i| \cdot||Te_i||\leq C < \infty$$ for some $C\in \mathbb{R}$ and $$n_k\leq ||Tx_{n_k}|| \leq |\; ||Tx||-||Tx_{n_k}|| \;|\leq ||Tx-Tx_{n_k}|| \leq \sum\limits_{k=1}^N |a_i-a_{i,n_k}| \cdot||Te_i||$$ which implies that $|a_{j,n_k}-a_j|$ can't converge towards $0$ for at least one $j\in \lbrace 1,...,N\rbrace$ because the left side diverges. Contradiction.
Hence $V$ can't be finite dimensional since $B_1(0) \supset \lbrace y \in V \; | \;||y||= 1\rbrace$ can't be compact in $V$.
• How did you know that $V$ is compact iff V is finite dimensional? I know that a way to prove this is to use that the isomorphism between $V$ and $K^n$ is continuous, but this is what we have to prove. – user370967 Feb 18 '18 at 18:38