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I am currently self-studying and came across the following theorem:

Suppose that a sequence of functions $\phi_n$ converges uniformly to $0$ on $[a,b]$. Now suppose we have a sequence of functions $f_n$ and a function $f$ on $[a,b]$ such that $$|f_n(x)-f(x)|\le \phi_n(x)$$ for all $x$ in $[a,b]$. It is stated that $f_n$ converges uniformly to $f$ on $[a,b]$.

However, I'm having difficulty understanding why this is obvious. Could someone please explain this to me? Any help would be greatly appreciated.

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  • $\begingroup$ (Uniformly) $f_n→f$ iff $f_n−f→0$; and if $|g_n|≤ϕ_n$ and $ϕ_n→0$ then $g_n→0$. $\endgroup$ Commented May 4 at 21:00

2 Answers 2

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$\phi_n$ converges uniformly to $0$ on $[a, b]$ means for all $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that

$$n \geq N \implies \lvert \phi_n(x) \rvert < \epsilon$$

for all $x \in [a, b]$. Observe

$$\lvert f_n(x) - f(x) \rvert \leq \phi_n(x) \implies \lvert f_n(x) - f(x) \rvert \leq \lvert \phi_n(x) \rvert$$

Therefore

$$n \geq N \implies \lvert f_n(x) - f(x) \rvert < \epsilon$$

which is the definition of uniform convergence of $f_n$ to $f$.

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For any function $g:[a,b]\to\Bbb R$, define $$\|g\|:=\sup\{|g(x)|:x\in[a,b]\}\in[0,\infty].$$ By hypothesis, $\|f_n-f\|\le \|\phi_n\|$ and $\|\phi_n\|$ converges to $0$, hence so does $\|f_n-f\|$.

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