Supremum of a sequence

Given is the set $$D := \{x \in \mathbb{R}\ |\ x^2-4x < 5\}$$.

Is $$sup\ D = \{5,-1\}$$ or $$sup\ D = 5$$?

Defining one variable function on specific interval and taking supremum of it

I got a course in my undergraduate education 4 years ago. So actually I can’t remember the details about mathematica. I’m trying to solve one thing about 1 week. Here is my question:

f is a continuos and bounded function on [a,b]. Define

blank”>{x,t\in[a,b]&space;and&space;|t-x|<\delta}|f(x)-f(t)|” title=”w(f,\delta)=sup_{x,t\in[a,b] and |t-x|<\delta}|f(x)-f(t)|” />

I want to write this function on mathematica and checking while delta going zero, is w will be going with it. I replaced supremum with MaxValue function. But I didn’t write when x,t in [a,b] take f(x)-f(t). Because f is a single variable and I dont know how can I put condition when t-x < delta.

These are my problems. After defining this function I want to see on graph how it changes with different delta’s

Linear programs with strict inequalities and supremum objectives

Linear programming can solve only problems with weak inequalities, such as “maximize $$c x$$ such that $$A x \leq b$$“. This makes sense, since problems with strict inequality often do not have a solution. For example “maximize $$x$$ such that $$x<5$$” does not have a solution.

But suppose we are interested in finding supremum instead of maximum. In this case, the above program does have a solution – the supremum is $$5$$.

Given a linear program with strict inequalities and a supremum or infimum objective, is it possible to solve it by reduction to a standard linear program?

Question about infimum and supremum in a positive cone

A “lattice cone” $$C$$ is a cone such that $$x\wedge y$$ and $$x\vee y$$ exist for $$x,y\in C$$.

But how can a postive cone contain both the supremum and the infimum? Since the relation

$$x\wedge y=x+y- x\vee y$$ would imply that $$x\wedge y\in C-C\neq C$$.

Maybe I’m confused with the definitions, would use some help here.

Why essential supremum would fail if essup is replaced by sup?

I know that $$\Vert f\Vert=esssup_{x}f(x)=0$$ if and only if $$f(x)=0$$ a.e., but why this property would fail if ess sup is replaced by sup? The sup means that $$\Vert f\Vert =\sup_{x} \vert f(x)\vert$$

A linear mapping from a finite dimensional space always attains its supremum over the basis of that space

Let $$f$$ be a linear mapping from the finite dimensional normed space $$X$$ to another normed space $$Y$$.

The proof for “A linear map from a finite-dimensional space is always continuous”, as given here, uses the fact that such a functional is continuous if and only if it is bounded.

Defining $$(e_1,\cdots, e_n)$$ as a basis for $$X$$, somewhere in the proof it is taken for granted that $$M:= \sup_i\{||f(e_i)||\}$$ does exist because $$X$$ is finite dimensional.

Now maybe I am missing something here, but as far as I remember, theorems about a mapping attaining its sup and inf on f.d. spaces assume its continuity, which is what this theorem is trying to establish.

Any hints where to look for? How is it guaranteed that $$M$$ as defined above is bounded?

Proving equivalence between $\epsilon$ based & $lub$ definition of supremum.

Based on $$\epsilon$$ have a new definition of supremum:

Let there be a nonempty set $$X$$ with supremum $$s$$, then $$X\cap(s – \epsilon, s]\ne \emptyset, \,\, \forall \epsilon\gt 0$$.

The conventional definition is given by:

Let $$X$$ be a nonempty set of real numbers. The number $$s$$ is called the supremum of $$X$$ if $$s$$ is an upper bound of $$X$$ and $$s \le y$$ for every upper bound of $$X$$.

Let, the conventional definition be denoted by ‘Def. 1’, while the new definition by ‘Def. 2’.

Have two questions below. I need help in attempting them.

Q. 1 : Need show that the two definitions are equivalent by proving the following two conditional statements:

(i) If $$s = sup(X)$$, as given by Defn. 1, then $$s$$ is the supremum, as given by Defn. 2. Here, assume that Defn. 1 holds, and use this assumption to prove that Defn. 2 holds.

Let $$s’$$ is supremum as per Defn. 2. Also, the relation between the magnitudes of $$s,s’$$ is unknown, & need be established.

$$s$$ will have set $$X$$ elements in the range $$(s-\epsilon, s]$$ if $$s-s’ \lt \epsilon$$, by the below proof:

Let us assume that $$s-s’ \ne 0$$, let $$s-s’=k.\epsilon, k\lt 1$$, then $$s = s’+k.\epsilon \implies s -\epsilon = s’+(k-1).\epsilon \implies s -\epsilon \lt s’$$.

$$s-\epsilon\lt s’\implies \exists x \in X: X\cap (s – \epsilon, s]\ne \emptyset$$.
But, Def. 2 can take any $$\epsilon\gt 0$$ to ensure $$\exists x \in X: X\cap (s’ – \epsilon, s’]\ne \emptyset$$.
So, if Def. 1 is to have ability to take any $$\epsilon\gt 0$$, need the lower bound of $$(s – \epsilon, s]$$ to equal at least to $$s’ – \epsilon$$.
But, $$s – \epsilon= s’+(k-1)\epsilon \ge s- \epsilon, \forall k, 0\lt k\lt 1$$.
So, the only possible value is $$k=0$$ to have the lower bound of $$(s – \epsilon, s]$$ equal to $$s’ – \epsilon$$.

But, by this cannot impose any restriction on the upper bound $$s$$ (of Def. 1) to equal $$s’$$ (of Def. 2).

(ii) If $$s = sup(X)$$, as given by Defn. 2, then $$s$$ is the supremum, as given by Defn. 1. Here, assume that Defn. 2 holds, and use this assumption to prove that Defn. 1 holds.

Let us modify for consistency with part (i) sake, $$s$$ replaced by $$s’$$.

If Defn. 2 holds, then the upper bound of the interval is bounded by $$s’$$, which is also the last element that can possibly be (if, $$s’\in X$$) in $$X$$. For Defn. 1 to hold, the upper bound must then be the same as the upper bound of Defn. 2, i.e. $$s’$$.

Q. 2: What is the practical significance of showing that these two definitions are logically equivalent?

The step (i) of showing that if Defn. 1 holds, then Defn. 2 holds, leads to having the lower bound of $$(s – \epsilon, s]=s’ – \epsilon$$.

The step (ii) of showing that if Defn. 2 holds, then Defn. 1 holds, leads to having the upper bound of $$(s – \epsilon, s]=s’$$per bound implies the max. value is $$s=1$$ at $$n= \infty$$. All possible values of $$n$$ are covered in the interval $$(s-\epsilon,s]$$ with $$n=\infty$$ at the upper bound.

On the supremum of the real parts of the zeros of the Riemann zeta function

Define $$S$$ to be the set of the real parts of the nontrivial zeros of the Riemann zeta function. The Prime Number Theorem entails that $$\zeta(s)\neq 0$$ for $$\Re(s)\geq 1$$. Does this mean the supremum of $$S$$ is less than $$1$$ ?

Expectation of supremum of random walk

I am stuck with the following problem: Let $$X$$ be a general discrete Random Walk and $$G$$ a geometric distributed random variable with parameter $$1-a$$. I would like to deduce from

(i) $$E(\sup_{n < G} X_n) < \infty$$

that

(ii) $$E(\sup_{n \geq 0} a^n X_n) < \infty$$

holds. I probably show that (i) $$\geq$$ (ii) and I assume, that the representation $$a^n = P(G > n)$$ is somewhat helpful.

I am grateful for any tips!

Essential Supremum and Supremum inside Expectation

Suppose that $$\{Z_i\}_{i \in I}$$ are a family of densities in $$L^2(\Omega,\mathcal{F},\mathbb{P})$$, and $$X=L^2(\Omega,\mathcal{F},\mathbb{P})$$. When is it true that $$\sup_{i \in I} \mathbb{E}\left[Z_i\cdot (X- \mathbb{E}[Z_i\cdot X|\mathcal{G}])^2 \right] = \sup_{i \in I} \mathbb{E}\left[Z_i\cdot (X- \operatorname{esssup}_{i \in I}\mathbb{E}[Z_i\cdot X|\mathcal{G}])^2 \right] ?$$

I’ve seen similar questions asked, but I haven’t come across something like this on here yet.