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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$$\alpha$/ ||F|| hence infimum must be greater than or equal to . But I stuck in other way pleaz help... thank you
Recall that $$||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 $\frac{\alpha}{||F||}$.
