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Prove that: A linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuous.

My attempt:

I know that $T$ will be continuous if and only if $ \lim\limits_{x \to a}Tx=Ta$. In other words, The goal is to show that for any $\epsilon>0,$ there exists $\delta>0$ such that $|x-a|<\delta$ implies $|Tx-Ta|<\epsilon$.

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    Can you show that it is enough to show that it is bounded? – 2017-02-15
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    First prove there exists a number $L > 0$ such that $\|Ax\| \leq L \|x\|$ for all $x$. – 2017-02-15

1 Answers 1

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Hint 1 : Consider any basis $(e_1,\cdots\,e_n)$ of $\mathbb{R}^n$ and try to obtain an upper bound for $\Vert T(x)\Vert$ in terms of $\Vert T(e_1)\vert$, ..., $\Vert T(e_1)\Vert$.

Hint 2 : For all $x\in\mathbb{R}^n$, there exist $x_1,\cdots,x_n\in\mathbb{R}$ such that $x=\sum_{i=1}^nx_ie_i$. Since $T$ is linear : $T(x)=\sum_{i=1}^nx_iT(e_i)$ and by triangle inequality :

$$\Vert T(x)\Vert\le\sum_{i=1}^n\vert x_i\vert\Vert T(e_i)\Vert$$

Here $\Vert y\Vert$ denotes the norm of a vector $y\in\mathbb{R}^m$ (how this norm is defined doesn't matter).

Let $M=\max\{\Vert T(e_i)\Vert;\,1\le i\le n\}$. Then we have :

$$\forall x\in\mathbb{R}^n,\,\Vert T(x)\Vert\le M\Vert x\Vert_1$$

where $\Vert \,\Vert_1$ denotes the norm defined on $\mathbb{R^n}$ by the sum of the absolute values of the coordinates relative to the basis $(e_1,\cdots,e_n)$.

It doesn't remain much work to reach the conclusion.

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    Not sure what that means but here is another thing that might help... From a previous problem I know that for any $a \in \mathbb{R}^n$ , $Tx-Ta=T(x-a)\leq M|x-a|$ – 2017-02-15
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    That basically solves the problem... just take $\delta =\epsilon / M$ – 2017-02-15
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    ...and don't forget to take the norm of the vector $T(x)-T(a)$. Comparison between a vector and a real number is not allowed ... – 2017-02-15