Problem statement

Let $g \in C[-1,1]$. Consider the set

$A=\{f \in C[-1,1] : f(x)\leq g(x), \space \forall x \in [-1,1]\}$.

$a)$ Prove that on $(C[0,1],d_ {\infty})$, $A^ {\circ}=\{f \in C[-1,1], f(x)<g(x), \space \forall x \in [-1,1]\}$.

$b)$ Prove that on $(C[0,1],d_1)$, $A$ is a closed set.

My attempt at a solution.

For $a)$ all I could think of is: If I want to prove that $A^ {\circ}=\{f \in C[-1,1], f(x)<g(x), \space \forall x \in [-1,1]\}$ on $(C[0,1],d_ { \infty})$, I should show the double inclusion of these two sets.

$S=\{f \in C[-1,1], f(x)<g(x), \space \forall x \in [-1,1]\} \subset A^ {\circ}$.

Let $f \in S$. I don't know how to justify this but there is $\delta>0: g(x)-\delta <f(x) <g(x)+\delta$ $\forall x \in [-1,1]$. How could I find $\epsilon>0$ such that if $h \in B_{d_ {\infty}}(f,\epsilon) \implies h(x)\leq g(x) \space \forall x \in [-1,1]?$

And for the inclusion $A^ {\circ} \subset S=\{f \in C[-1,1], f(x)<g(x), \space \forall x \in [-1,1]\}$, suppose $f \in A^ {\circ}$, then there is $\epsilon>0 :B_{d_ {\infty}}(f,\epsilon) \subset A$, i.e., $sup_{x \in [0,1]}|f(x)-h(x)|<\epsilon \implies h \in A$. I don't understand how can I deduce from here that $f(x)\leq g(x) \space \forall x \in [-1,1]$.

As for part $b)$, in this one I am totally lost, I remind that $d_1(f,g)=\int_0^1 |f(x)-g(x)|dx$. Could it be easier to prove $A^c$ is open?

I would appreciate some guidance, suggestions or hints on how could I solve this exercise.

  • $\begingroup$ For part (b), what is $B$? Is it $A$? $\endgroup$ – user99914 Dec 11 '13 at 3:40
  • $\begingroup$ Yes, sorry, in the exercise it was called $B$, but I've decided to change it to $A$ and I forgot to change that part. I've corrected it. $\endgroup$ – user100106 Dec 11 '13 at 3:45


For part (a), to show $S\subset A^0$, pick $f\in S$. Then by definition $f(x) < g(x)$ for all $x\in [0,1]$. Then there is $\epsilon $ independent of $x$ such that $f(x) + \epsilon < g(x)$ for all $x\in [0,1]$ (Why?).

For $A^0 \subset S$, then $f\in A^0$. Then by definition there is $\epsilon>0$ such that $B(f, \epsilon) \subset A$. In particular, the function $f(x) + \epsilon/2 $ is in $A$.

For part (b), we can show directly that $A$ is closed under $d_1$. Let $f_n \in A$ and $f_n \to f$ in the metric $d_1$. Assume the contrary that $f\notin A$. Then by definition there is $x_0\in [0,1]$ such that $f(x_0) > g(x_0)$. As $f, g$ are continuous, there is $\delta_1, \delta_2$ such that $f(x) > g(x) + \delta_1$ for all $|x-x_0|<\delta_2$. Now consider

$$ d_1(f_n , f) = \int^1_0 |f_n - f| \geq \int_{x_0 - \delta_2}^{x_0+ \delta_2} |f_n - f|\ ,$$

now try to use the fact $f_n \in A$ to find a positive lower bound for the above expression.

  • $\begingroup$ John: I understood what you did to show that $A^{\circ} \subset S$, however, for the other parts I didn't get what you did: to show $S \subset A^{\circ}$, I understand there has to be $\epsilon$ such that $f(x)+\epsilon <g(x)$ for all $x \in [0,1]$ because if not, there would be an $x_0$ in the interval : $f(x_0)\geq g(x_0)$, contradicting the fact that $f \in S$. I suppose you are suggesting me to consider this same $\epsilon$ and see that $B(f,\epsilon) \subset A$. I don't see why this $\epsilon$ works. $\endgroup$ – user100106 Dec 11 '13 at 4:50
  • $\begingroup$ Consider the function $f(x) = 1/x$ on $(0,\infty)$. Then $f(x)>0$ for all $x$. But you cannot find a fixed $\epsilon$ such that $f(x) > \epsilon $ for all $x$. Having such an $\epsilon$ is not so trivial. $\endgroup$ – user99914 Dec 11 '13 at 4:52
  • $\begingroup$ And for part $b)$, you are trying to conclude that if $f \not \in A$, then $f_n \not \to f$, which is absurd. I don't understand why the integral $\int_{x_0 - \delta_2}^{x_0+ \delta_2} |f_n - f|$ would help me to arrive to an absurd. Sorry if I am asking too obvious things here. $\endgroup$ – user100106 Dec 11 '13 at 4:52
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    $\begingroup$ For (b), I have given more hints in the answer, please see if it is okay. $\endgroup$ – user99914 Dec 11 '13 at 4:56
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    $\begingroup$ Hmmm, ok, you gave an example where $f$ is not defined on a compact set; in the original problem, $f$ attains a maximum and $g$ attains a minimum, I suppose that the existence of $\epsilon$ could be ensure by this fact. $\endgroup$ – user100106 Dec 11 '13 at 4:57

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