How to find infimum of below set

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$$\alpha/ ||F|| hence infimum must be greater than or equal to . But I stuck in other way pleaz help... thank you 1 Answer Recall that$$||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||}$$as ||x_i||=1 and ||F(x_i)||\to ||F||>0. Thus, the infimum is less than or equal to \frac{\alpha}{||F||}. • Why this$$\left|\left|\frac{\alpha}{||F||}x_i\right|\right |$\$ is in the set. Note it not satisfy the required condition May 11, 2017 at 19:15
• Please reply if i miss something May 11, 2017 at 19:19
• Ooops, you are right. I'll update my answer when I figure out how to fix this. May 11, 2017 at 19:24
• I have fixed my solution. May 11, 2017 at 19:27
• No problem! If my answer was what you are looking for, please feel free to accept it. May 11, 2017 at 19:34