推 bineapple :好方法!十分感謝~~ 10/16 23:29
※ 引述《bineapple (パイナップル)》之銘言：
: f:R→R
: If there is at least one point x_0 in R where f is continuous on it,
: and f(x+y)=f(x)+f(y) for all x and y in R, show that f(x)=ax for some a.
: 我已經會證明當x_0不是0的情況了
: 想請問有高手能提示一下是0時的情況嗎??
: 謝謝
Since f(x+y)=f(x)+f(y) for all x and y in R, we have
f(rx) = rf(x) for all x in Ｒ for all r in Ｑ.
Let a = f(1). So f(r) = ar for all r in Ｑ. Since Ｑ is dense in Ｒ,
and f is continuous on x , we have f(x_0) = a x .
0 0
Let and α be in Ｒ. For h in Ｒ and α-x_0-h is rational,
we have |f(α) - aα| = |f(α+ x_0 + h) - f(x_0 + h) - aα|
= |a(α+ x_0 + h) - f(x_0 + h) - aα| = |a(x_0 + h) - f(x_0 + h)|.
Take limit on h tending to 0, |f(α) - aα| = 0.
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