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 !!!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.