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Let $f:\mathbb{R}^n\to \mathbb{R}^m$ be a linear function, and let $e_1, e_2, \dots, e_n$ be the standard basis vectors for $\mathbb{R}^n$. Let $w_1, w_1, \dots, w_m\in\mathbb{R}^m$ be the vectors defined by $w_i=f(e_i), 1\le i \le n$. I want to prove that $\|f(x)\|\le c\|x\|$, for all $x\in\mathbb{R}^n$. Also prove that it is Lipschitz.

My proof:

Let $x\in\mathbb{R}^n$, then $x=\alpha_1 e_1+\dots+\alpha_n e_n$ for some scalars $\alpha_i$, $1\le i \le n$. Then $f(x) = f(\alpha_1 e_1+\dots+\alpha_n e_n)=\alpha_1 f(e_1)+\dots+\alpha_n f(e_n)=\alpha_1 w_1 + \dots + \alpha_n w_n$, so that $\|f(x)\|=\|\alpha_1 w_1 + \dots + \alpha_n w_n\|$ $=|(\alpha_1e_1+\dots+\alpha_ne_n)\cdot(w_1+\dots+w_n)|$ $\le \|x\|\| w_1 + \dots + w_n\|\le \|x\|(\|w_1\|+\dots+\|w_n\|)=c\|x\|$.

Moreover, this function is Lipschitz: let $x,y\in\mathbb{R}^n$, and consider $\|f(x)-f(y)\|\le \|f(x-y)\|\le c\|x-y\|$.

Please let me know if you think my proof is fine. I'm not so sure about the part where I take the dot product of vectors with different bases.

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    $\begingroup$ Taking the dot product of $e$'s and $w$'s doesn't make sense here since the $e$'s are in $\Bbb R^n$ and the $w$'s are in $\Bbb R^m$, where presumably $m\neq n$ in general. $\endgroup$ – Dave May 16 '17 at 3:09
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As mentioned by @Dave, the part you were doubtful about is indeed wrong (also if $n=m$. Think of the case where $w_1$ is, say, $e_2$), but you were on the right path, and can fix it with Cauchy-Schwarz. Let $c_i = \|w_i\|$.

Suppose that $x=\sum_{i=1}^n \alpha_i e_i$. Then $\|x\|=\sqrt{\sum_{i=1}^n |\alpha_i|^2}$. Now
$$\|f(x)\|=\|\sum_{i=1}^n \alpha_i w_i\| \le \sum_{i=1}^n |\alpha_i|c_i \le \sqrt{\sum_{i=1}^n |\alpha_i|^2}\sqrt{\sum_{i=1}^n c_i^2}=c\|x\|\mbox{ where }c=\sqrt{\sum_{i=1}^n c_i^2}. $$ The first inequality is by the triangle inequality, and the second by Cauchy-Schwarz.

The proof of the Lipschitz property is correct, with one minor comment: by linearity, $f(x)-f(y)=f(x-y)$, so you can use an equality instead of the first inequality.

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  • $\begingroup$ Can you please clarify what would happen if $w_1 = e_2$? $\endgroup$ – sequence May 16 '17 at 14:33
  • $\begingroup$ Let $f$ be such that $w_1 = f(e_1) = e_2,w_2= f(e_2) = -e_1$. Choose $x= e_1+e_2$. Then $f(x) = -e_1+e_2$, so $\|f(x)\|=\sqrt{2}$. But $x \cdot (w_1+w_2) = (e_1+e_2)\cdot (-e_1+e_2) = 0$. $\endgroup$ – Fnacool May 18 '17 at 20:38

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