Assume that $\pi:(V_1,||.||_1)\to V_2$ is a surjective linear map from a normed vector space into a vector space. How to find a norm $||.||_2$ on $V_2$ using this map?


Hint: If $\pi : V_1 \to V_2$ is a linear surjection, $\bar{\pi}: V_1/\ker \pi \to V_2$ (defined by $\bar{\pi}(x+\ker \pi) = \pi(x)$) will be a bijection. Now how can you use $\lVert \cdot \rVert_1$ to define a norm on $V_1/\ker \pi$?

  • $\begingroup$ In this case one can define a norm $||.||$ on $V_1/ker \pi$ as $||x+ker \pi||=||x||_1$ and then easily define a norm $||.||_2$ on $V_2$ as $||w||_2=||\pi(x+ker \pi)||_2=||x||_1$. $\endgroup$ – Majid Jun 17 '17 at 18:24
  • $\begingroup$ @Majid Nope, that won't work. Your norm isn't well defined. Suppose $x,y \in V_1$ with $x -y \in \ker \pi$. Then $x+ \ker \pi = y + \ker \pi$ but it isn't necessarily true that $\lVert x \rVert_1 = \lVert y \rVert_1$. $\endgroup$ – Demophilus Jun 17 '17 at 18:28
  • $\begingroup$ @Demophilis yes you are right. what about this: Let $\bar x=x+ker \pi$. Then $|| \bar x||=inf \{||x||_1 : \bar x=x+ker\pi\}$? $\endgroup$ – Majid Jun 17 '17 at 18:51
  • $\begingroup$ @Majid That is indeed a norm, which isn't too hard to verify. $\endgroup$ – Demophilus Jun 17 '17 at 18:54
  • $\begingroup$ Now my question is that why people easily do not use this solution? Why they bother themselves defining the unit balls on the vector spaces and then defining the norm using the unit ball on $V_2$? It does not mean that something must be wrong with this way? $\endgroup$ – Majid Jun 17 '17 at 18:57

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