Editing Chebyshev polynomials (section)

===Minimal {{math|∞}}-norm===
For any given {{math|''n'' ≥ 1}}, among the polynomials of degree {{mvar|n}} with leading coefficient 1 ([[monic polynomial|monic]] polynomials):
<math display="block">f(x) = \frac{1}{\,2^{n-1}\,}\,T_n(x)</math>
is the one of which the maximal absolute value on the interval {{closed-closed|−1, 1}} is minimal.

This maximal absolute value is:
<math display="block">\frac1{2^{n-1}}</math>
and {{math|{{abs|''f''(''x'')}}}} reaches this maximum exactly {{math|''n'' + 1}} times at:
<math display="block">x = \cos \frac{k\pi}{n}\quad\text{for }0 \le k \le n.</math>

{{Math proof | proof =
Let's assume that {{math|''w<sub>n</sub>''(''x'')}} is a polynomial of degree {{mvar|n}} with leading coefficient 1 with maximal absolute value on the interval {{closed-closed|−1, 1}} less than {{math|1&thinsp;/&thinsp;2<sup>''n''&nbsp;−&thinsp;1</sup>}}.

Define
<math display="block">f_n(x) = \frac{1}{\,2^{n-1}\,}\,T_n(x) - w_n(x)</math>

Because at extreme points of {{mvar|T<sub>n</sub>}} we have
<math display="block">\begin{align}
|w_n(x)| &< \left|\frac1{2^{n-1}}T_n(x)\right| \\
f_n(x) &> 0 \qquad \text{ for }~ x = \cos \frac{2k\pi}{n} ~&&\text{ where } 0 \le 2k \le n \\
f_n(x) &< 0 \qquad \text{ for }~ x = \cos \frac{(2k + 1)\pi}{n} ~&&\text{ where } 0 \le 2k + 1 \le n
\end{align}</math>

From the [[intermediate value theorem]], {{math|''f<sub>n</sub>''(''x'')}} has at least {{mvar|n}} roots.  However, this is impossible, as {{math|''f<sub>n</sub>''(''x'')}} is a polynomial of degree {{math|''n'' − 1}}, so the [[fundamental theorem of algebra]] implies it has at most {{math|''n'' − 1}} roots.
}}

====Remark====
By the [[equioscillation theorem]], among all the polynomials of degree {{math|≤&thinsp;''n''}}, the polynomial {{mvar|f}} minimizes {{math|{{norm|&thinsp;''f''&thinsp;}}<sub>∞</sub>}} on {{closed-closed|−1, 1}} [[if and only if]] there are {{math|''n'' + 2}} points {{math|−1 ≤ ''x''<sub>0</sub> < ''x''<sub>1</sub> < ⋯ < ''x''<sub>''n'' + 1</sub> ≤ 1}} such that {{math|1={{abs|&thinsp;''f''(''x<sub>i</sub>'')}} = {{norm|&thinsp;''f''&thinsp;}}<sub>∞</sub>}}.

Of course, the null polynomial on the interval {{closed-closed|−1, 1}} can be approximated by itself and minimizes the {{math|∞}}-norm.

Above, however, {{math|{{abs|&thinsp;''f''&thinsp;}}}} reaches its maximum only {{math|''n'' + 1}} times because we are searching for the best polynomial of degree {{math|''n'' ≥ 1}} (therefore the theorem evoked previously cannot be used).