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.

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)}$ $

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.

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> }