Post by A-gine[1;33;40mIt's easy to see that sup |fn(x)| = sup fn(x)[m
[1;33;40m |x|≦1 [0,1][m
^^^^^^^^^^^^^^^^^^^^^^^^^^
[1;37;40m這個看不太出來,麻煩大大提示>""<[m
For x in (0,1), d/dx fn(x) = (1+nx^2 - 2nx^2)/(1+nx^2)^2
= (1-nx^2)/(1+nx^2)^2
If fn has a local maximum at x, 1=nx^2 => fn(x) = 1/(2√n)
fn(0) = 0 < fn(x) and fn(1) = 1/(1+n) ≦ 1/(2√n) = fn(x);
hence, sup |fn(x)| = 1/(2√n).
Since |fn| is contionous and [-1,1] is compact, there exist pn in [-1,1]
such that |fn(pn)| = sup |fn|.
Since |fn| is even function, WLOG we may assume pn≧0;
hence, sup|fn| = sup |fn| = sup fn
[-1,1] [0,1] [0,1]
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[m
[1;36m弄清π是無理數這件事可能是根本沒有實際用處的[m
[1;36m但是如果我們能弄清楚那麼肯定就不能容忍不去設法把它弄清楚[m
[1;36m ──E.C.Titchmarsh[m
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