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Assume $a \in \mathbb{R},g(a)$ is the maximum of $|x^2-ax|$ in $[0, 1]$, what is $a$ when $g(a)$ attains its minimum?

My solution is direct and therefore lengthy. I guess there is a simpler or systematic method of solving similar problems?

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2 Answers 2

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Well since $f(x)=|x^2-ax|$ has maximum at $p={a\over 2}$ or $f(1) = |1-a|$. So: $$g(a) ={1\over 2}\Big({a^2\over 4} + |1-a|+\Big|{a^2\over 4}-|1-a|\Big|\Big)$$ Now if $a\geq 1$ we get:

$$ g(a) = {1\over 8}\Big((a -2)^2+a^2+4a-4\Big) =a^2/4 \geq 1/4$$

If $a<1$ we get:

$$ g(a) = {1\over 8}\Big((a -2)^2+|a^2+4a-4|\Big) \geq |1-a| $$

Here we used triangle inequality $|a^2-4a+4|+|-a^2-4a+4|\geq |8-8a|$.

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  • $\begingroup$ I think your solution is wrong because the form of g(a) is incorrect. $\endgroup$
    – tmpbin
    Nov 19, 2017 at 23:46
  • $\begingroup$ And what is the correct form? $\endgroup$
    – nonuser
    Nov 20, 2017 at 9:59
  • $\begingroup$ a/2 should be a^2/4 $\endgroup$
    – tmpbin
    Nov 20, 2017 at 10:04
  • $\begingroup$ Ahh, yes. My fault. Thanks. $\endgroup$
    – nonuser
    Nov 20, 2017 at 10:05
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Hint:

For the global maximum, you can't avoid studying the expression at both endpoints of the interval and find local maxima, if any.

  • value at $x=0$: $0$;

  • value at $x=1$: $|1-a|$;

  • the local maxima of the absolute value are the local extrema of the expression, provided it occurs in the given interval. We have a minimum at $(\frac a2,-\frac{a^2}4)$, corresponding to a maximum of the absolute value at $(\frac a2,\frac{a^2}4)$.

The plot below summarizes this:

enter image description here

I doubt there is a much shorter way.

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