# 5 Minute Daily Practice - Test Taking - Grades 4 - 8 by Jacqueline B. Glasthal

By Jacqueline B. Glasthal

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**Example text**

Solution. By definition of f ∗ , f ∗ (y) = sup(y T x − f (x)). x ∗ If y ∈ dom f , then the affine function h(x) = y T x−f ∗ (y), minorizes f . Conversely, if h(x) = aT x + b minorizes f , then a ∈ dom f ∗ and f ∗ (a) ≤ −b. The set of all affine functions that minorize f is therefore exactly equal to the set of all functions h(x) = y T x + c where y ∈ dom f ∗ , c ≤ −f ∗ (y). 28, f (x) = sup y∈dom f ∗ (y T x − f ∗ (y)) = f ∗∗ (y). 40 Gradient and Hessian of conjugate function. Suppose f : Rn → R is convex and twice continuously differentiable.

This shows that dmp is convex, since it is the maximum of 2n linear functions of (p, q). Let’s now identify the (or a) subset C that maximizes prob(p, C) − prob(q, C) = i∈C (pi − qi ). The solution is C = {i ∈ {1, . . , n} | pi > qi }. Let’s show this. The indices for which pi = qi clearly don’t matter, so we will ignore them, and assume without loss of generality that for each index, p> qi or pi < qi . Now consider any other subset C. If there is an element k in C but not C, then by adding k to C we increase prob(p, C) − prob(q, C) by pk − qk > 0, so C could not have been optimal.

What can you say about convexity or concavity of g? Solution. g is concave. Its hypograph is hypo g = = = {(y, t) | t ≤ g(y)} {(y, t) | f (t) ≤ f (g(y))} {(y, t) | f (t) ≤ y)} 0 1 = 1 0 (because f is increasing) epi f. For differentiable g, f , we can also prove the result as follows. Differentiate g(f (x)) = x once to get g (f (x)) = 1/f (x). so g is increasing. Differentiate again to get g (f (x)) = − f (x) , f (x)3 so g is concave. 4 [RV73, page 15] Show that a continuous function f : Rn → R is convex if and only if for every line segment, its average value on the segment is less than or equal to the average of its values at the endpoints of the segment: For every x, y ∈ Rn , 1 0 f (x + λ(y − x)) dλ ≤ f (x) + f (y) .