I have a function $$\delta:\mathbb{R}\rightarrow [0,1].$$

We obtain this funtion pointwise as follows: For each point $y\in\mathbb{R}$, $\delta(y)$ is a real number in $[0,1]$. More explicitely, $\delta(y)$ can be any number in $[0,1]$ given a specific $y$.

Once $\delta$ is specified for all $y\in\mathbb{R}$, then $\delta$ will have as many elements as ${\mathbb{R}}$. As we are free to chose any $\delta(y)\in[0,1]$ given a specific $y$, $\delta$ can be obtained in uncountably many ways.

The set $\Delta$ is composed of any possible construction of $\delta$.

Below is a discrete example when $y$ and $\delta$ are defined as sets. In the original question $y\in\mathbb{R}$ and $\delta$ was a continuos function on real numbers.

Discrete example: Let $y:=\{0,1,2\}$ and $\delta(y)\in\{0.1,0.5\}$, Then






and we have $\Delta=\{\Delta_1,\Delta_2...,\Delta_8\}$

I must prove that the set $\Delta$ of all possible $\delta(y)\in \Delta$ is convex and compact.

Convexity: For any given $\alpha\in[0,1]$

$$\delta^{'}(y)=\alpha\delta(y)+(1-\alpha)\delta(y)$$ is also a valid function in $\Delta$, therefore $\Delta$ is a convex set.

Compactness: If I can show that for each open cover of $\Delta$ there exists a subcover, then I am done but I dont know how to show it or if there is something simpler.

I will be very grateful for any help,

Thanks in advance.

  • $\begingroup$ Say again, what is $\Delta$? $\endgroup$ – Did Nov 26 '12 at 12:59
  • $\begingroup$ $\Delta$ is the set of all possible functions $\delta(y)$ due to any arbitrary definition/choice of $u(y)$. For example assume, we are dealing with the discrete case, and we have $u(1)$, $u(2)$ and $u(3)$, $y\in\{1,3\}$. Then I can have $\delta(1)=0.4$, $\delta(2)=0.8$, $\delta(3)=0.3$ and have $\Delta_1=\{0.4,0.8,0.3\}$ or I could have $\delta(2)=0.1$, $\delta(2)=0.1$, $\delta(3)=0.2$ or etc.. and $\Delta:=\{\Delta_1,\Delta_2...\}$ $\endgroup$ – Seyhmus Güngören Nov 26 '12 at 13:06
  • $\begingroup$ Still lost... Is $\Delta$ equal to $[0,1]^\mathbb R$? $\endgroup$ – Did Nov 26 '12 at 13:09
  • $\begingroup$ Each element of $\Delta$ is $[0,1]^{\mathbb{R}}$, because $\delta(y)\in[0,1]^{\mathbb{R}}$. However one can define $u(y)$ as he/she wishes, and therefore the set $\Delta$ contains uncountably many members that are $\in[0,1]^{\mathbb{R}}$ $\endgroup$ – Seyhmus Güngören Nov 26 '12 at 13:13
  • $\begingroup$ You seem to be saying that $\Delta=\{\delta(y)\mid y\in Y\}\subseteq[0,1]^\mathbb R$ for some unspecified set $Y$, and that each $\delta(y)$ is more or less any element of $[0,1]^\mathbb R$ one wants. Then, why should one expect $\Delta$ to be convex (or measurable)? $\endgroup$ – Did Nov 26 '12 at 13:21

In the end, it seems that $\Delta$ is simply $\Delta=[0,1]^\mathbb R$ the set of all functions defined on $\mathbb R$ with values in $[0,1]$. Then the convexity of $\Delta$ is trivial since any product of convex sets is convex and $\Delta$ is a product of the convex set $[0,1]$. Is this set $\Delta$ compact? For the product topology, this is Tychonoff theorem.

But now I note that later on in the question, you declare that every $\delta$ in $\Delta$ is a continuous function... Thus, $\Delta$ would be $\Delta=C^0(\mathbb R,[0,1])$. Then the convexity of $\Delta$ is trivial. Is this set $\Delta$ compact? This could depend on the topology you put on $\Delta$ (pointwise convergence? uniform convergence?) but in general the answer shall be no.

Note Two confusions seem to plague your understanding:

  • First, functions are not numbers. You cannot at the same time pretend that For each point $y\in\mathbb{R}$, $\delta(y)$ is a real number in $[0,1]$ and that $\delta(y)$ is a function defined on $\mathbb R$. If the first assertion holds, then $\delta$ is a function and $\delta(y)$ is a number. Please watch out for this confusion when you write down your definitions.

  • Second, one does not check convexity the way you check it: if $\delta$ is in a set $\Delta$, then $\delta'=\alpha\delta+(1-\alpha)\delta$ is also in $\Delta$, always! Simply because $\delta'=\delta$. Convexity asks something else, namely that $\alpha\delta_1+(1-\alpha)\delta_2$ is in $\Delta$, for every $\alpha$ in $[0,1]$ and every $\delta_1$ and $\delta_2$ in $\Delta$.

Finally, note that when I declare that the convexity of any product of convex sets is trivial, I mean it. Please write the definitions down and you should see that this is obvious and that the only hypothesis you need is that $[0,1]$ (the target set) is convex. Ditto for the convexity of $C^0(\mathbb R,[0,1])$.

Edit It appears now that $\Delta$ would be the set of continuous nondecreasing functions $\delta$ defined on $\mathbb R$ such that $0\leqslant\delta\leqslant1$, $\delta(y)=0$ for every $y$ small enough $y$ and $\delta(y)=1$ for every $y$ large enough. Then, convexity is still trivial and compactness still dubious (consider the functions $(\delta_t)_{t\in\mathbb R}$ in $\Delta$ defined by $\delta_0:y\mapsto\max\{0,\min\{y,1\}\}$ and, for every $t$, $\delta_t:y\mapsto\delta_0(y-t)$).

  • $\begingroup$ In the paper he was mentioning that $\Delta$ was compact with respect to the infinity norm. Does it say something to you? $\endgroup$ – Seyhmus Güngören Nov 27 '12 at 21:38
  • $\begingroup$ Sure, this is the topology of the uniform convergence--then we are probably in the second case of my answer, namely $\Delta=C^0(\mathbb R,[0,1])$. But why should this space be compact when the source set $\mathbb R$ is not, eludes me. (What is the source?) $\endgroup$ – Did Nov 27 '12 at 21:45
  • $\begingroup$ I am sorry, I know one source it was in the matrix movie)) here it is arxiv.org/pdf/0707.2926.pdf on page 3, after equation 2.6. $\endgroup$ – Seyhmus Güngören Nov 27 '12 at 23:11
  • $\begingroup$ The assertion just after equation 2.6 that $\|\delta\|_\infty=\max\limits_{y\in\mathbb R}\delta(y)$ needs some serious justification. Without further hypothesis, this is wrong, consider $\delta(y)=y^2/(y^2+1)$. $\endgroup$ – Did Nov 27 '12 at 23:19
  • $\begingroup$ The highest restriction on $\delta$ is that it is $0$ from $-\infty$ to some real number $y_l$, then increasing from $0$ to $1$ from $y_l$ to $y_u$ and for $y>y_u$, it is one. Is this restriction useful? I want to consider your example but I dont know how) $\endgroup$ – Seyhmus Güngören Nov 27 '12 at 23:40

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