- #1
epr1990
- 26
- 0
Homework Statement
Let [itex]\left[a,b\right][/itex] be a closed bounded interval, [itex]f : [a,b] \rightarrow \textbf{R}[/itex] be bounded, and let [itex]g : [a,b] \rightarrow \textbf{R}[/itex] be continuous with [itex]g\left(a\right)=g\left(b\right)=0[/itex]. Let [itex]f_{n}[/itex] be a uniformly bounded sequence of functions on [itex]\left[a,b\right][/itex]. Prove that if [itex]f_{n}\rightarrow f[/itex] uniformly on all closed intervals [itex]\left[c,d\right]\subset\left(a,b\right)[/itex], then [itex]f_{n}g\rightarrow fg[/itex] uniformly on [itex]\left[a,b\right][/itex]
Homework Equations
f is bounded on [a,b]:
[itex](\exists M_{f}\in\textbf{R}) \ni : (\left|f(x)\right| \leq M_{f}) (\forall x\in\left[a,b\right])[/itex]
g is continuous on [a,b]:
[itex](\forall x\in\left[a,b\right]) (\forall\epsilon>0) (\exists\delta>0) (\forall y\in\left[a,b\right]) \ni : (\left|x-y\right|<\delta \Rightarrow \left|f\left(x\right) - f\left(y\right)\right|<\epsilon) [/itex]
with [itex]g\left(a\right)=g\left(b\right)=0[/itex]
f_n is a bounded sequence of functions on [a,b]:
[itex](\exists M_{f_{n}}\in\textbf{R}) \ni : (\left|f(x)\right| \leq M_{f_{n}}) (\forall x\in\left[a,b\right]) (\forall n\in \textbf{N})[/itex]
f_n converges uniformly to f on all closed subsets [c,d] of (a,b):
[itex](\forall \epsilon >0)(\exists N \in\textbf{N}) \ni : (n\geq N \Rightarrow |f_{n}(x) - f(x)| < \epsilon) (\forall x\in[c,d])(\forall [c,d] \subset (a,b))[/itex]
The Attempt at a Solution
Proving that [itex]f_{n}g\rightarrow fg[/itex] uniformly on [itex]\left[c,d\right][/itex] on all closed intervals [itex]\left[c,d\right]\subset\left(a,b\right)[/itex], is fairly trivial. Given that [itex]\left[a,b\right][/itex] is a closed bounded interval, with g continuous on [itex]\left[a,b\right][/itex], the extreme value theorem holds. So, g is bounded on [itex]\left[a,b\right][/itex] by [itex]M_{g}=sup_{x\in[a,b]}g(x)[/itex], and since [itex]\left[c,d\right]\subset\left(a,b\right)[/itex] is a closed subset, then [itex]|g(x)|\leq M_{g}[/itex], [itex]\forall x \in [c,d][/itex]. Now, if [itex]f_{n}\rightarrow f[/itex] uniformly on all closed intervals [itex]\left[c,d\right]\subset\left(a,b\right)[/itex], then fixing [itex]\epsilon > 0[/itex], we can choose an [itex]N\in\textbf{N}[/itex], so that, [itex]\forall n\geq N[/itex], [itex]|f_{n}(x) - f(x)| < \epsilon / M_{g}[/itex]. Thus, [itex]|f_{n}(x)g(x) - f(x)g(x)| < \epsilon[/itex].
However, I don't see anyway to extend this to [a,b]. In the previous exercise in the book, it asked to prove that if f_n(x) is bounded on a set E for each n and f_n converges uniformly to f on E, then f_n is uniformly bounded on E, and f is bounded on E. I did that, and that proof was easy, but I don't see where it could apply here. I don't think the converse is true in general, so the only thing that I could gain from that is that f is bounded on [c,d], but I already have that since it is bounded on [a,b]. Clearly, I need to use the fact that f_n is uniformly bounded on [a,b] and f is bounded on [a,b] and probably use the continuity of g on [a,b] instead of the restriction to [c,d] that I used here. On top of this, I don't see a single way in which g(a)=g(b)=0 would apply. Any suggestions?
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