Let $F$ be a non-zero continuous linear map from $X$ to $Y$ (both normed linear space). Let $\alpha$ be positive, then show that infimum of the set $\inf\{\|x\|:\|F(x)\|=\alpha\}$ is $\alpha/\|F\|$.

I have done one inequality as every member in set must satisfy $\|F(x)\| = \alpha$ therefore

$\alpha$ $\leq $ ||F|| ||x|| hence ||x|| $\geq$$\alpha$/ ||F|| hence infimum must be greater than or equal to . But I stuck in other way pleaz help... thank you


1 Answer 1


Recall that $$||F||=\sup_{||x||=1}||F(x)||.$$ Take a sequence $x_i$ in the unit ball such that $||F(x_i)||\to ||F||$. It follows that $$\left|\left|F\left(\frac{\alpha}{||F(x_i)||}x_i\right)\right|\right|=\frac{\alpha}{||F(x_i)||}||F(x_i)||=\alpha.$$ Also, for all $i$, we have $$\left|\left|\frac{\alpha}{||F(x_i)||}x_i\right|\right|=\frac{\alpha}{||F(x_i)||}\to\frac{\alpha}{||F||}$$ as $||x_i||=1$ and $||F(x_i)||\to ||F||>0$. Thus, the infimum is less than or equal to $\frac{\alpha}{||F||}$.

  • $\begingroup$ Why this $$\left|\left|\frac{\alpha}{||F||}x_i\right|\right | $$ is in the set. Note it not satisfy the required condition $\endgroup$ May 11, 2017 at 19:15
  • $\begingroup$ Please reply if i miss something $\endgroup$ May 11, 2017 at 19:19
  • $\begingroup$ Ooops, you are right. I'll update my answer when I figure out how to fix this. $\endgroup$ May 11, 2017 at 19:24
  • $\begingroup$ I have fixed my solution. $\endgroup$ May 11, 2017 at 19:27
  • $\begingroup$ No problem! If my answer was what you are looking for, please feel free to accept it. $\endgroup$ May 11, 2017 at 19:34

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