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f ′ n(x) = ne −n2 x 2
(1 − 2n 2 x 2 ) x + n = 1/n √ 2
x − n = −1/n √ 2
sup
x∈R
|fn(x) − f(x)| = sup |fn(x)| ≥ |f(x
x∈R
± n )| = 1
√ e
2 −2 .
n → +∞
R 2 fn(x) 0
0 {x : |x| ≥ ε} ε > 0 |fn(x)| = |fn(−x)|
sup |fn(x) − f(x)| = sup |fn(x)|.
|x|≥ε
|x|≥ε
n x + n , x − n ∈] − ε, ε[ |f(ε)| > |f(x)| |x| ≥ ε
|F |
x + n , x − n ] − ∞, x − n [ ]x + n , +∞[
sup |fn(x) − f(x)| = |fn(ε)|,
|x|≥ε
0
R
fn(0) = 0 x = 0 |fn(x)| n → ∞
f(x) = 0 R2 fn
x + n = 1/n x− n = −1/2
|fn(x ± n ) − f(x)| = 1/2
0
0
fn(0) = 0 x = 0 fn(x) 1 n → +∞ fn
[0, 1] f f(0) = 0
f(x) = 1 x ∈]0, 1] [0, 1]
f
[ε, 1] ε > 0
sup |fn(x) − f(x)| = sup |fn(x) − 1| = sup
1
[ε,1]
[ε,1]
[ε,1] 1 + nx
=
1
1 + nε
0 n → +∞
x2 − x ≤ 1/2 x ∈ [0, 1] |fn(x)| ≤ 1/2n → 0 x ∈ [0, 1]