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Let $C([a,b])$ be the set of continuous real valued function defined on the interval $[a,b]$, for ${-\infty<a<b<\infty}$. Define a subset $A \subset C([a,b])$ is open if, for every $f \in A$, there exists some $\epsilon_{f}>0$ such that all $g\in C([a,b])$ which satisfy $||f-g||:=\max({|g(x)-f(x)|:x\in[a,b]})< \epsilon_{f}$ are also in $A$.

(a) Show that the collection of open sets in $c([a,b])$ is a topology.

(b) Show that $E:C([a,b])\rightarrow \mathbb{R}$ given by $E(f):= \int_{a}^{b} x f(x)dx$ is continuous.

In first part of the problem the space should follow $3$ axioms $\phi$ and $X$ (set) should lie in the space and the union and intersection should lie in the space. I am facing issue in formally defining function and taking forward.

In the second part, any hint would be helpful !!!

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  • $\begingroup$ I am not sure I understand what you want. Can you more clearly state the 3 axioms ? are you sure that union and intersection are on the same footing ? $\endgroup$ – lmsteffan Sep 28 '14 at 15:37
  • $\begingroup$ So your question is about part (b), not part (a)? Or are you having problems with part (a) as well? $\endgroup$ – Paul Sundheim Sep 28 '14 at 15:43
  • $\begingroup$ Thanks for the comment. I am facing problem with 'a' as well. The union and intersection do not come under same footing, they will come separately. $\endgroup$ – vikram Sep 28 '14 at 15:53
  • $\begingroup$ For part (a) you could just prove that the sets $ B(f,\epsilon_f) = \left{ g \in C \left( \left[ a,b \right] \right) \text{:} \| f-g \| < \epsilon_f \right} $ form a basis for the topology on $C\left(\left[a,b\right]\right)$. $\endgroup$ – Yeldarbskich Sep 28 '14 at 16:02
  • $\begingroup$ This will tell me that the set is open, as the ball of radius $\epsilon_f$ lies in it. But how do I check the axioms especially the first one where $\phi$ and X lie in the space. $\endgroup$ – vikram Sep 28 '14 at 20:13
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For part a you use the fact that every metric space is also a topological space. The metric you have is called total variation and the topological space created by this metric is an open ball of radius $\epsilon_f$ for each $f$. Since $A$ results from the collection of topological spaces for every $f\in A$, it is also a topology, satisfying its axioms.

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  • $\begingroup$ Don't I need to check the axioms or just defining the set as open is sufficient. Just bit confused here. $\endgroup$ – vikram Sep 28 '14 at 20:15
  • $\begingroup$ I was able to solve the first part given the above explanation. Any hint for the second part? $\endgroup$ – vikram Sep 29 '14 at 12:48
  • $\begingroup$ @vikram For part $b$ I dont have very concrete statements. I think $\delta-\epsilon$ definition of continuity solves the problem. $g(x)=f(x)+\delta$ and $|E(g)-E(f)|<\epsilon$. $\endgroup$ – Seyhmus Güngören Sep 30 '14 at 22:53

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