[分析] 一致性連續一問

Uniform continuity 的定義是:
f(x) is uniformly continuous over I if
for any ε>0 there exists a δ>0 such that
whenever |x-y|<δ, x,y in I, then |f(x)-f(y)|<ε.

我們知道 f(x)=x^2 continuous on R, 因此當然也
continuous on [0,+∞). 但它不是 uniformly continuous,
其關鍵在於 uniformly continuous 定義中的 δ 與 x, y
無關, 而普通的 pointwise continuity 是 fix x, 而取的
δ 可以與 x 有關.

從定義來看, 要證明不是 uniformly continuous, 就是
證明:
For some ε0>0 such that for any δ>0,
there exists a sequence of (x_n,y_n)
such that
|y_n-x_n|<δ for all n,
　 but |f(y_n)-f(x_n)|>ε0 for infinitly many n.

考慮 y_n = x_n + δ' where 0<δ'<δ, 例如取 δ'=δ/2.
則
|f(y_n)-f(x_n)| = |(x_n+δ')^2-x_n^2| = |2δ'x_n+δ'^2| > (2x_n)δ'
當 x_n↑+∞ 時, |f(y_n)-f(x_n)|→+∞.
因此, for fixed ε0>0, for any δ>0, we have a sequence of (x_n,y_n)
such that
　 |y_n-x_n|<δ for all n, but |f(y_n)-f(x_n)|>ε0 for sufficiently large n.
這證明了 f(x)=x^2 在 [0,+∞) 不是 uniformly continuous.

f(x) = x^2

Let x_n=sqrt(n+1), y_n=sqrt(n)

x_n - y_n ->0 as n -> infinite

but f(x_n) - f(y_n) = n+1 - n = 1 -/-> 0

"the area between the funtion and the x axis
is not uniform at the same distance"

f(x)=x^2
take [0,1] and [1,2]

[0,1]
∫f(x)dx=(1/3)x^3=1/3
(1/3) / 1= 1/3

[1,2]
∫f(x)dx=(1/3)x^3=7/3
(7/3) / 1= 7/3

1/3 is not equal to 7/3
f(x) is not uniform continuous