From Wiki

Every linear function on a finite-dimensional space is continuous.

I was wondering what the domain and codomain of such linear function are?

Are they any two topological vector spaces (not necessarily the same), as along as the domain is finite-dimensional? Can the codomain be a different normed space (and may not be finite-dimensional)?

I asked this because I saw elsewhere the same statement except the domain is a finite-dimensional normed space, and am also not sure if the codomain can be a different normed space (and may not be finite-dimensional).

Thanks and regards!

  • 2
    $\begingroup$ The codomain can we any normed space, finite dimensional or not. In fact, you can check that if $E$ is a vector space of dimension $n$ over the field of real numbers and $F$ any normed space, and $T\colon E\to F$ a linear map then $T(E)$ has a dimension $\leq n$. $T$ is continuous since we can take a basis $v_1,\ldots,v_k$ of $R(T)$ and we write for $x\in E$ $T(x)=\sum_{i=1}^ka_i(x)v_i$ and check that each $a_i$ is linear and continuous. $\endgroup$ Feb 24, 2012 at 20:17
  • $\begingroup$ Continuity of a linear map is equivalent to boundedness: math.stackexchange.com/questions/86451/… $\endgroup$
    – dls
    Feb 24, 2012 at 20:52
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    $\begingroup$ To emphasize the key point: a "topological vector space" is more than a vector space endowed with a topology. It is a vector space endowed with a topology making the vector space operations continuous. When $V$ is finite dimensional over $\mathbb{R}$ or $\mathbb{C}$, then there is a unique "topological vector space" topology on $V$ (and any norm on $V$ induces it). This uniqueness means that people often talk about "finite dimensional spaces" without explicitly mentioning the topology (as that Wiki page does). But a specific topology is always intended. The result is not true in general. $\endgroup$ Feb 24, 2012 at 23:06
  • $\begingroup$ For example, there are lots of nonconstant linear functions from $\mathbb{R}$ with its usual vector space structure, to $\mathbb{R}$ with its usual vector space structure (these are the maps $x \mapsto ax$, with $a$ nonzero). But if you endow the domain with the usual topology, and the codomain with the discrete topology, then none of these nonconstant linear maps are continuous. So, to get the result you want, it's crucial that specific topologies are understood on the spaces (and indeed, the properties of these topologies are used heavily in any proof of that result). $\endgroup$ Feb 24, 2012 at 23:09
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    $\begingroup$ @leslie: Your uniqueness statement has an error: there is a unique nondiscrete (Hausdorff) topological vector space structure on any finite-dimensional real or complex vector space. (It goes through for finite-dim. spaces over p-adic fields too.) $\endgroup$
    – KCd
    Feb 25, 2012 at 9:27

3 Answers 3


The result we can show is the following:

Let $E$ and $F$ two topological vector spaces, where $E$ is Hausdorff, and $T\colon E\to F$ a linear map. If $E$ is finite dimensional, then $T$ is continuous.

First, if $(e_1,\ldots,e_n)$ is a basis of $E$, then any set of $n+1$ vectors of $T(E)$ is linearly dependent, so $T(E)$ has a dimension $\leqslant n$. Let $k$ be the dimension of $T(E)$, and $(v_1,\ldots,v_k)$ a basis of this space. We can write for any $x\in E$: $T(x)=\sum_{i=1}^ka_i(x)v_i$ and since $v_i$ is a basis each $a_i$ is linear. We have to show that each map $T_i\colon E\to F$, $T_i(x)=:a_i(x)v_i$ is continuous.

Added: the map $x\mapsto a_i(x)$ is well-defined because $(v_1,\ldots,v_k)$ is a basis. In particular, it takes finite values.

By definition of a topology on a topological vector space we only have to show that the map $x\mapsto a_i(x)$ is continuous. To do that, we use the fact that a finite dimensional Hausdorff topological vector space can be equipped with a norm which gives the same topology (in fact it is the unique one), namely put $$N\left(\sum_{j=1}^n\alpha_j e_j\right):=\sum_{j=1}^n|\alpha_j|$$ Now the continuity is easy to check: denoting $x=\sum_{j=1}^nx_je_j$ and $y=\sum_{j=1}^ny_je_j$ $$|a_i(x)-a_i(y)|\leqslant \sum_{j=1}^n|a_i((x_j-y_j)e_j)|=\sum_{j=1}^n|x_j-y_j|\cdot |a_i(e_j)| \leqslant N(x-y)\sum_{j=1}^n|a_i(e_j)|,$$ since $|x_j-y_j|\leqslant N(x-y)$ for all $1\leqslant j\leqslant n$.

  • $\begingroup$ Thanks! (1) "a finite dimensional topological space can be equipped with a norm which gives the same topology", do you mean topological vector space instead of topological space? (2) Why is the last inequality? $\endgroup$
    – Tim
    Feb 25, 2012 at 1:06
  • $\begingroup$ Yes I meant topological vector space, I will fix it and give more details for the last inequality. $\endgroup$ Feb 25, 2012 at 8:49
  • $\begingroup$ Thanks! (1) In the definition of norm $N$, I suppose $x_j$ is $e_j$? (2) Regarding your detail explanation about the equality asked in my last comment, is there an inequality for a general normed space, like $||x .\times y|| \leq ||x|| \cdot ||y||$ for some elementwise multiplication $.\times$ between two vectors? In order to make "element" to have sense, do we have to make $.\times$ independent of basis? $\endgroup$
    – Tim
    Feb 25, 2012 at 12:54
  • $\begingroup$ How do you see that $a_i(e_j)$ cannot be infinite? $\endgroup$
    – PPR
    Mar 8, 2015 at 18:51
  • 1
    $\begingroup$ @PPR Please see the edit. $\endgroup$ Mar 8, 2015 at 20:04

There is a nice write-up of a proof of a simpler version of this (for normed vector spaces) on Wikipedia. Let me reproduce the proof given on Wikipedia:

If $T:X\to Y$ is a linear map where $X,Y$ are normed vector spaces and $X$ has finite dimension then $T$ is continuous.

Proof: Let $e_1,\dots, e_n$ be a basis of $X$. Then for $x \in X$ we have

$$ \|Tx\|_Y = \|T \sum_i \alpha_i e_i\|_Y = \|\sum_i \alpha_i Te_i \|_Y \le \sum_i |\alpha_i|\|Te_i\|_Y$$

Let $\varepsilon >0$ be given and let $M = \max_i \|Te_i\|_Y$. Let $\delta = {\varepsilon \over M}$. Then for $x$ with $\|x\| < \delta$ we have

$$ \|Tx\|_Y \le \sum_i |\alpha_i|\|Te_i\|_Y \le M \sum_i |\alpha_i| < \varepsilon$$

hence $T$ is continuous.

  • 2
    $\begingroup$ I think you missed a constant while you bound $\sum_{i}|a_i|$ by $||x||$. $\endgroup$
    – perlman
    Nov 1, 2017 at 21:08
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    $\begingroup$ Why does $\lVert x\rVert< \delta$ mean that $\sum_i \lvert \alpha_i\rvert$ is small? A priori you don't know the the norm of $x$ is related to its basis coefficients right? $\endgroup$
    – John
    Nov 22, 2017 at 12:14
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    $\begingroup$ Yes, you do know that, because all norms are equivalent on a finite dimensional space. $\endgroup$
    – user370967
    Feb 22, 2018 at 22:25

The special case of a linear transformations $A: \mathbb{R}^n \to \mathbb{R}^n$ being continuous leads nicely into the definition and existence of the operator norm of a matrix as proved in these notes.

To summarise that argument, if we identify $M_n(\mathbb{R})$ with $\mathbb{R^{n^2}}$, and suppose that $v \in \mathbb{R}^n$ has co-ordinates $v_j$, then by properties of the Euclidean and sup norm on $\mathbb{R}^n$ we have:

$\begin{align}||Av|| &\leq \sqrt{n} \,||Av||_{\sup} \\&= \sqrt{n}\max_i\bigg|\sum_{j}a_{ij}\,v_j\bigg|\\&\leq \sqrt{n}\max_i \sum_{j}|a_{ij}\,v_j|\\&\leq \sqrt{n} \max_i n\big(\max_j|a_{ij} v_j|\big)\\&=n\sqrt{n} \max_i \big(\max_j |a_{ij}| \max_j |v_j|\big)\\&= n\sqrt{n}\max_{i,j}|a_{ij}|||v||_{\sup}\\&\leq n \sqrt{n} \max_{i,j}|a_{ij}|||v|| \end{align}$

$\Rightarrow ||Av|| \leq C ||v||$ where $C = n\sqrt{n}\displaystyle\max_{i,j}|a_{ij}|$ is independent of $v$

So if $\varepsilon>0$ is given, choose $\delta = \dfrac{\varepsilon}{C}$ and for $v, w \in \mathbb{R}^n$ with $||v-w||< \delta$ consider

$||Av - Aw || = ||A(v-w) || \leq C ||v-w || < \delta C= \varepsilon$ from which we conclude that $A$ is uniformly continuous.


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