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x ∈ [0, +∞[
fn
f(x) =
∞
1
e−xt dt
1 + t2 R [0, +∞[
[0, +∞[
|f(x, y) − fn(x, y)| =
∞
1
e−xt dt −
1 + t2 n
1
e−xt
dt
1 + t2 =
∞
n
e−xt π
dt ≤ − arctan n,
1 + t2 2
fn(x) ≤ f(x) x
eα < 1 α < 0
sup |f(x) − fn(x)| ≤
x∈[0,+∞[
π
− arctan n,
2
[0, +∞[
fn :
R 2 → R fn(x, y) =
2 n (x + y)
1 + n2 n (x 2 + y 2 )
fn f(x, y) = 0
R 2 fn(0, 0) = 0 limn→∞ fn(0, 0) = 0 (x, y) = (0, 0)
|fn(x) − f(x)| = |fn(x)| ≤ 1
n
|x + y|
(x 2 + y 2 )
n → ∞ 1 +
n2 n (x 2 + y 2 ) > n2 n (x 2 + y 2 )
1
1 + n2 n (x 2 + y 2 ) <
1
n2 n (x 2 + y 2 ) .
fn
f fn
sup
(x,y)∈R2 |fn(x, y) − f(x, y)| = sup
(x,y)∈R2 |fn(x, y)| = sup |fn(ρ cos θ, ρ sin θ)|
ρ≥0
θ∈[0,2π]
= sup
ρ≥0
θ∈[0,2π]
2nρ 1 + n2n | cos θ + sin θ|
ρ2 | cos θ + sin θ| ≤ √ 2 θ ∈ [0, 2π]
θ1 = π/4 θ2 = 5π/4
sup |fn(ρ cos θ, ρ sin θ)| = sup
ρ≥0
ρ≥0
θ∈[0,2π]
2 n√ 2ρ
1 + n2 n ρ 2 = √ 2 sup
ρ≥0
2 n ρ
1 + n2 n ρ 2
g(θ) := cos θ + sin θ [0, 2π]
0 = − sin θ + cos θ θ = π/2, 3/2π tan θ = 1 θ1 = π/4 θ2 = 5π/4
|g(θ1)| = |g(θ2)| = √ 2 > 1 = |g(π/2)| = |g(3π/2)| = |g(0)| = |g(2π)|.