Example of convex functions fulfilling a (strange) lower bound

I am reading a preliminary version of a paper which focuses on some minimization problems connected to a class of integral functionals. Reading the assumptions of one of the theorems I cannot convince myself that a “concrete” example exists.

Question. Does there exist a function $$f \colon \mathbb R^N \to [0,+\infty)$$ such that

1. $$f$$ is convex;
2. $$f(\lambda x) = \lambda f(x)$$ for any $$\lambda >0$$ and $$\forall x \in \mathbb R^N$$;
3. there are $$a>0, b \ge 0$$ and $$\gamma \in \mathbb R^N$$ such that $$a|x| \le f(x) + \langle \gamma, x \rangle + b$$ for any $$x \in \mathbb R^N$$?

I have some problems in finding a function satisfying the three points… It does not have to be smooth, still I do not see an example.

upper bound of consecutive integers which are not coprime with n!

Is there any research on getting upper bound of consecutive integers which less than $$n!$$ and NOT coprime with $$n!$$?

Easy to see that lower bound $$>=n$$, (example that $$n= odd prime$$). Does upper bound $$<2n$$ strictly?

Note that the discussion is in range of $$.

Given least upper bound $\alpha$ for $\{\ f(x) : x \in [a,b] \ \}$, $\forall \epsilon > 0 \ \exists x$ s.t. $\alpha – f(x) < \epsilon$

I can’t figure out how all of this follows. Taken from Ch.8 of Spivak’s Calculus.

If $$\alpha$$ is the least upper bound of $$\{\ f(x) : x \in [a,b] \ \}$$ then, $$\forall \epsilon > 0 \ \exists x\in [a,b] \ \ \ \ \ \ \ \alpha – f(x) < \epsilon$$ This, in turn, means that $$\frac{1}{\epsilon} < \frac{1}{\alpha – f(x)}$$

Upper bound for the sup of the first derivative

Given $$f : \mathbb{R} \to \mathbb{R}$$ which is twice differentiable and that

$$M_0 = \sup_{x \in \mathbb{R} } |f(x)|$$

$$M_1 = \sup_{x \in \mathbb{R} } |f'(x)|$$

$$M_2 = \sup_{x \in \mathbb{R} } |f”(x)|$$

prove that $$M_1^2 \leq 2 M_0 M_2$$ ?

I think a good approach is Taylor series but for what values !

Quantitative bound on irrational rotation recurrence time

Given an irrational $$a$$, the sequence $$b_n := na$$ is dense and equidistributed in $$\mathbb S^1$$ where we view $$\mathbb S^1$$ as $$[0, 1]$$ with its endpoints identified.

Given a point $$p$$ in $$\mathbb S^1$$, can we obtain a quantitative upper bound (that can depend on $$a, p, e$$) on the smallest $$n$$ such that $$na$$ is in $$B_e (p)$$?

Proof of a lower bound of the recurrence relation (the CLRS’s 4.6-2 exercise)

I am trying to find a solution to the ex. 4.6-2 of the Introduction to Algorithms by Cormen, Leiserson, Rivest, Stein (the third edition). It requires, for recurrence relations $$T(n)=aT(n/b)+f(n)$$ where $$a\geq 1, b > 1$$, $$n$$ is an exact power of $$b$$ and $$f(n)$$ is an asymptotically positive function, to prove that if $$f(n) = \Theta(n^{\log_ba}lg^{k}n)$$, where $$k\geq0$$, then $$T(n)=\Theta(n^{\log_ba}lg^{k+1}n)$$.

Since $$T(n) = n^{\log_ba} + g(n)$$ where $$g(n) = \sum_{j=0}^{\log_b n – 1} a^{j}f(n/b^{j})$$, I decided to consider the $$g(n)$$ function at the first. And I have shown $$g(n) = O(n^{\log_ba}lg^{k+1}n)$$ already (I think so).

But the doing the proof $$g(n) = \Omega(n^{\log_ba}lg^{k+1}n)$$ became a challenge for me.

Below is my research (with the simplified assumption $$k$$ is an integer). By condtition, there is such constant $$c$$ that

$$g(n) \geq$$

$$c \sum_{j=0}^{\log_b n – 1} a^{j}(n/b^{j})^{log_ba}log^{k}(n/b^{j}) =$$

$$cn^{\log_ba}\sum_{j=0}^{\log_b n – 1} log^{k}(n/b^{j}) =$$

$$cn^{\log_ba}\sum_{j=0}^{\log_b n – 1}(logn – logb^{j})^{k} =$$

$$cn^{\log_ba}\sum_{j=0}^{\log_b n – 1}\sum_{i=0}^{k} {k \choose i}log^{k-i}n(-logb^{j})^{i} =$$

$$cn^{\log_ba}log^{k}n\sum_{j=0}^{\log_b n – 1}\sum_{i=0}^{k} {k \choose i}(-logb^{j}/logn)^{i} =$$

$$cn^{\log_ba}log^{k}n \biggl(log_bn + \sum_{j=0}^{\log_b n – 1}\sum_{i=1}^{k} {k \choose i}(-logb^{j}/logn)^{i} \biggr) \geq$$

$$c’n^{\log_ba}log^{k+1}n – cn^{\log_ba}log^{k}n\sum_{j=0}^{\log_b n – 1}\sum_{i=1}^{k} {k \choose i}(logb^{j}/logn)^{i} =$$

$$A(n) – B(n) = \Theta(n^{\log_ba}lg^{k+1}n) – B(n)$$

Actually I am stuck with it. I can not show that $$B(n)$$ grows slower than $$A(n)$$. For instance, since $$(logb^{j}/logn)^{i} \lt 1$$ we are able to enhance our $$\geq$$ condition by the substitution $$B(n)$$ to some fucntion $$B'(n)$$ with the sums of binominal coefficients only. But then finally $$B'(n)$$ has $$n^{\log_ba}log^{k+1}n$$.

So how to prove $$g(n) = \Omega(n^{\log_ba}lg^{k+1}n)$$ ?

Bound on the mutual information between a product of correlated random variables

Let $$G$$ be a finite group.

Suppose the random variables $$X_1,\dots,X_N$$ are sampled uniformly at random from $$G$$. Let $$Y_1,\dots,Y_N$$ be random variables where $$Y_i$$ is correlated with $$X_i$$ and sampled according to some unknown distribution.

Given a bound on the mutual information $$I(X_k:Y_k) \leq \epsilon_k$$ for all $$k$$, what is a good upper bound on $$I(X_1\dots X_N: Y_1\dots Y_N)$$? I.e., the product of the two sets of random variables.

I believe a bound like $$I(X_1\dots X_N: Y_1\dots Y_N) \leq C\prod\epsilon_k$$ for some $$C$$ might exist but have had no luck in proving it.

Bound on volume of Minkowski Difference

$$A-B:=\{c:B+c\subseteq A\}.$$

I think that if $$A-B$$ is not the universe, then $$vol(A-B)\leq vol(A)$$ (If $$B$$ has one point then the inequality is immediate, adding more points to $$B$$ further reduces $$vol(A-B)$$ by set theoretic consideration.) Is there anything tighter?

Anti-concentration: upper bound for $P(\sup_{a \in \mathbb S_{n-1}}\sum_{i=1}^na_i^2Z_i^2 \ge \epsilon)$

Let $$\mathbb S_{n-1}$$ be the unit sphere in $$\mathbb R^n$$ and $$z_1,\ldots,z_n$$ be a i.i.d sample from $$\mathcal N(0, 1)$$.

Question

Given $$\epsilon > 0$$ (may be assumed to be very small), what is a reasonable upper bound for the tail probability $$P(\sup_{a \in \mathbb S_{n-1}}\sum_{i=1}^na_i^2z_i^2 \ge \epsilon)$$ ?

Observations

• Using ideas from this other answer (MO link), one can establish the non-uniform anti-concentration bound: $$P(\sum_{i=1}^na_i^2z_i^2 \le \epsilon) \le \sqrt{e\epsilon}$$ for all $$a \in \mathbb S_{n-1}$$.

• The uniform analogue is another story. May be one can use covering numbers ?

How to repeat bound controls based on clicking Add new in Asp.net mvc and Knockoutjs

I want to repeat a dropdownlist that is already bound using a Viewbag property and another textbox when a user click on Add Course. I have used asp.net mvc and knockout.js based on a tutorial i saw, but the tutorial does not exactly handle using bound controls, please how can i achieve this using asp.net mvc and knockout.js. Below is my code. Thanks

<table id="jobsplits">      <thead>         <tr>            <th>@Html.DisplayNameFor(m => m.selectedCourse.FirstOrDefault().FK_CourseId)</th>            <th>@Html.DisplayNameFor(m => m.selectedCourse.FirstOrDefault().CourseUnit)</th>            <th></th>         </tr>      </thead>      <tbody data-bind="foreach: courses">            @for (int i = 0; i < Model.selectedCourse.Count; i++)            {               <tr>                   <td>                        @Html.DropDownListFor(model => model.selectedCourse[i].FK_CourseId, new SelectList(ViewBag.Course, "Value", "Text", Model.FK_CourseId), "Select Course", new { @class = "form-control", data_bind = "value: courses" })                   </td>                   <td>                        @Html.TextBoxFor(model => model.selectedCourse[i].CourseUnit, new { htmlAttributes = new { @class = "form-control", @readonly = "readonly", data_bind = "value: courseUnit" } })                   </td>                   <td>                        <button type="button" data-bind="click: \$  root.removeCourse" class="btn delete">Delete</button>                   </td>               </tr>            }      </tbody> </table>  <div class="col-md-4">      <button data-bind="click: addCourse" type="button" class="btn">Add Course</button> </div> 

This is the script section

@section Scripts{ @Scripts.Render("~/bundles/knockout") <script>     function CourseAdd(course, courseUnit) {         var self = this;          self.course = course;         self.courseUnit = courseUnit;     }      function CourseRegViewModel() {         var self = this;          self.addCourse = function () {             self.courses.push(new CourseAdd(self.course, self.courseUnit));         }          self.courses = ko.observableArray([             new CourseAdd(self.course, self.courseUnit)         ]);          self.removeCourse = function (course) {             self.courses.remove(course)         }                 }      ko.applyBindings(new CourseRegViewModel()); </script> }