how to show the minimizing max has a solution.
confusing about how to approve it
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Hint to (4). By expanding we get $$\|\alpha f+g\|^2\geq\|g\|^2\iff \alpha^2\|f\|^2+2\alpha\langle f,g\rangle\geq0.$$ Now if $\langle f,g\rangle=0$ we have $\|\alpha f+g\|\geq\|g\|$ for any real $\alpha$.
If now $\langle f,g\rangle\neq0$ show that for $\alpha=-\dfrac{\langle f,g\rangle}{\|f\|^2}$ that $\|\alpha f+g\|<\|g\|$.
Michael Hoppe
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Unfortunately not; I know a about Chebyshev's approximation in context of functional analysis only. – Michael Hoppe Oct 16 '13 at 05:37