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I am trying to show from first principles that a linear map $T:X \rightarrow Y$ between finite dimensional vector spaces $X$ and $Y$ is Lipschitz. I'm rather stuck and would appreciate a hint how I might continue the argument I have constructed so far:

To show that $T$ is Lipschitz, it is necessary to show for $x,y \in U$ that $|T(x) - T(y)| \leq \lambda|x - y|$ for some $\lambda > 0$. If $u_1, \dots, u_n$ is a basis for $X$, there exists scalars $a^i, b^i$ such that $x = a^iu_i$ and $y=b^iu_i$. Therefore,

$$ |T(x - y)| \leq |T(a^iu_i)| + |T(b^iu_i)| = |a^i|\cdot|T(u_i)| + |b^i|\cdot|T(u_i)| \leq k \sum_{i=1}^n|T(u_i)| $$


$$k = \max_{j=1, \dots n}|a^j + b^j|$$

So, I've succeeded in finding a bounds of sorts for $T$ but I'm not sure how to translate the bound $|T(u_i)|$ to $u_i$ and other than showing something rather obvious so far, I'm not really sure that the above argument gets me a lot closer to my goal. How could I proceed with this?

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up vote 3 down vote accepted

There are a few ways you could proceed. For example, it is equivalent to showing that the following quantity is bounded: $\sup_{z\in X, \|z\| = 1}\|Tz\|$. Indeed, $\|T(x) - T(y)\|\leq \lambda\|x - y\|$ if and only if

$$ \frac{\|T(x) - T(y)\}}{\|x - y\|}\leq \lambda, $$

assuming that $x\neq y$. Now by linearity of $T$, the above inequality holds if and only if

$$ \left\|T\left(\frac{x - y}{\|x - y\|}\right)\right\|\leq \lambda. $$

Rewriting this with $z = x - y$, we get the claim.

To bound this, notice that $\|Tz\|\leq (|a_1| + \cdots + |a_n|)\max_{1\leq i\leq n}\|Te_i\|$, where $\{e_i\}$ is an orthonormal basis for $X$ and $z = \sum_{i=1}^n a_i e_i$. Since $e_1,\dots,e_n$ are orthonormal and $\|z\| = 1$, this puts a uniform (i.e. independent of $z$) bound on $|a_1|+\cdots+|a_n|$ (prove this).

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I should add that $\lambda$ in this case can be chosen (obviously) as $\sup_{z\in X, \|z\|=1}\|Tz\|$. In this case $\lambda$ is denoted by $\|T\|$ and is called the norm of $T$, and $T$ is said to be bounded. In general in a Banach space an operator is bounded if and only if it is continuous. So more so for finite dimensional normed linear spaces, every linear operator is continuous, and in fact smooth, which isn't difficult to show. – William Feb 14 '12 at 3:24

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