Doom of Doom
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Let )<C<\infty and a,b \in \mathbb{R}. Also let
Lip_{C}\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | \left|f(x) - f(y)\right| \leq C \left|x-y\right| \forall x,y \in \left[a,b\right]\right\}<br />.
Let \left(f_{n}\right) _{n \in \mathbb(N)} be a sequence of functions with f_{n} \in Lip_{C}\left(\left[a,b\right]\right) for all n.
i) Show that if \left(f_{n}\right) _{n \in \mathbb(N)} converges uniformly to a function f : \left[a,b\right]\rightarrow \mathbb{R}, then f \in Lip_{C}\left(\left[a,b\right]\right).
ii) Is Lip_{C}\left(\left[a,b\right]\right) a sub vector space of B\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | f bounded }?____________________________________
Useful formula:
The sequence converges uniformly, so:
For all \epsilon > 0, there exists N \in \mathbb{N} so that for all n,m > N, |f_{n}(x)-f_{m}(x)|<\epsilon for all x \in \left[a,b\right]\.
For (i), I honestly have no idea.
It seems that I can just say:
|f(x)-f(y)|
= |\lim_{n \to \infty} f_{n}(x) - \lim_{n \to \infty} f_{n}(y)|
= |\lim_{n \to \infty} (f_{n}(x) - f_{n}(y))|
= \lim_{n \to \infty} |(f_{n}(x) - f_{n}(y))| \leq C |x-y|
But I know that has to be wrong. It's not even using the uniformity.For (ii) I know the answer is of course not.
Let f(x):= Cx. Then 2 f \notin Lip_{C}\left(\left[a,b\right]\right), since
|2f(x)-2f(y)| = 2C|x-y| > C|x-y|.
Help on part (i) please?
Lip_{C}\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | \left|f(x) - f(y)\right| \leq C \left|x-y\right| \forall x,y \in \left[a,b\right]\right\}<br />.
Let \left(f_{n}\right) _{n \in \mathbb(N)} be a sequence of functions with f_{n} \in Lip_{C}\left(\left[a,b\right]\right) for all n.
i) Show that if \left(f_{n}\right) _{n \in \mathbb(N)} converges uniformly to a function f : \left[a,b\right]\rightarrow \mathbb{R}, then f \in Lip_{C}\left(\left[a,b\right]\right).
ii) Is Lip_{C}\left(\left[a,b\right]\right) a sub vector space of B\left(\left[a,b\right]\right) := \left\{f:\left[a,b\right]\rightarrow \mathbb{R} | f bounded }?____________________________________
Useful formula:
The sequence converges uniformly, so:
For all \epsilon > 0, there exists N \in \mathbb{N} so that for all n,m > N, |f_{n}(x)-f_{m}(x)|<\epsilon for all x \in \left[a,b\right]\.
For (i), I honestly have no idea.
It seems that I can just say:
|f(x)-f(y)|
= |\lim_{n \to \infty} f_{n}(x) - \lim_{n \to \infty} f_{n}(y)|
= |\lim_{n \to \infty} (f_{n}(x) - f_{n}(y))|
= \lim_{n \to \infty} |(f_{n}(x) - f_{n}(y))| \leq C |x-y|
But I know that has to be wrong. It's not even using the uniformity.For (ii) I know the answer is of course not.
Let f(x):= Cx. Then 2 f \notin Lip_{C}\left(\left[a,b\right]\right), since
|2f(x)-2f(y)| = 2C|x-y| > C|x-y|.
Help on part (i) please?