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Yes, if fₙ and gₙ are sequences of Real numbers such that converge uniformly to f and g, on A⊂ R respectively, then (fₙ gₙ) isconverges uniformly to (f g).
Suppose fn → f uniformly and gn → g uniformly
on A ⊆ R and that both f and g are bounded. Then (fₙ.gₙ) converges uniformly to (fg) on A.
Proof. Let ε > 0. Suppose M₁ > 0 is an upper bound for f (meaning |f(x)| ≤ M₁ for all x ∈ A) and that M₂ is an upper bound for g.
Since fn → f uniformly, there exists N₁ ∈ N such
that n ≥ N₁ implies |fn(x) − f(x)| < 1 for all x ∈ A. Equivalently, |fn(x)| < |f(x)| + 1 ≤ M₁ + 1 for all
x ∈ A.
In turn, since gn → g uniformly, there exists N₂ ∈ N such that n ≥ N₂ implies
|gn(x) − g(x)| < 2(M₂ + 1).
Finally, since fn → f uniformly, there exists N₃∈ N such that n ≥ N3 implies |fn(x) − f(x)| < 2M₂ .
Now, if N = max{N₁, N₂, N₃} and n ≥ N, we have
|fn(x)gn(x) − f(x)g(x)| = |fn(x)gn(x) − fn(x)g(x) + fn(x)g(x) − f(x)g(x)|
≤ |fn(x)gn(x) − fn(x)g(x)| + |fn(x)g(x) − f(x)g(x)|
= |fn(x)||gn(x) − g(x)| + |g(x)||fn(x) − f(x)|
≤ (M₁ + 1)|gn(x) − g(x)| + M₂|fn(x) − f(x)|
< (M₁ + 1) ε/2(M₁ + 1) + εM₂/2M₂
= ε for all x ∈ A So, we see that, indeed, (fngn) → (fg) uniformly on A. Hence proved.
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