Difference between revisions of "2000 USAMO Problems/Problem 1"

Latest revision as of 23:45, 8 July 2017

Problem

Call a real-valued function $f$ very convex if

$\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|$

holds for all real numbers $x$ and $y$ . Prove that no very convex function exists.

Solution

Solution 1

Let $y \ge x$ , and substitute $a = x, 2b = y-x$ . Then a function is very convex if $\frac{f(a) + f(a+2b)}{2} \ge f(a + b) + 2b$ , or rearranging,

$\left[\frac{f(a+2b)-f(a+b)}{b}\right]-\left[\frac{f(a+b)-f(a)}{b}\right] \ge 4$

Let $g(a) = \frac{f(a+b) - f(a)}{b}$ , which is the slope of the secant between $(a,f(a))(a+b,f(a+b))$ . Let $b$ be arbitrarily small; then it follows that $g(a+b) - g(a) > 4$ , $g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4$ . Summing these inequalities yields $g(a+kb)-g(a) > 4k$ . As $k \rightarrow \infty$ (but $b << \frac{1}{k}$ , so $bk < \epsilon$ is still arbitrarily small), we have $\lim_{k \rightarrow \infty} g(a+kb) - g(a) = g(a + \epsilon) - g(a) > \lim_{k \rightarrow \infty} 4k = \infty$ . This implies that in the vicinity of any $a$ , the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.

Solution 2

Suppose, for the sake of contradiction, that there exists a very convex function $f.$ Notice that $f(x)$ is convex if and only if $f(x) + C$ is convex, where $C$ is a constant. Thus, we may set $f(0) = 0$ for convenience.

Suppose that $f(1) = A$ and $f(-1) = B.$ By the very convex condition, $\frac{f(0) + f\left(2^{-n}\right)}{2} \ge f\left(2^{-(n+1)}\right) + \frac{1}{2^n}.$ A straightforward induction shows that: $f\left(2^{-n}\right) \le \frac{A - 2n}{2^n}$ for all nonnegative integers $n.$ Using a similar line of reasoning as above, $f\left(-2^{-n}\right) \le \frac{B - 2n}{2^n}.$ Therefore, for every nonnegative integer $n,$ $f\left(2^{-n}\right) + f\left(-2^{-n}\right) \le \frac{A+B-4n}{2^n}.$ Now, we choose $n$ large enough such that $n > \frac{A+B}{4} - 1.$ This is always possible because $A$ and $B$ are fixed for any particular function $f.$ It follows that: $f\left(2^{-n}\right) + f\left(-2^{-n}\right) < \frac{1}{2^{n-2}}.$ However, by the very convex condition, $f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.$ This is a contradiction. It follows that no very convex function exists.

@@ Line 1: / Line 1: @@
 == Problem ==
-Call a real-valued function <math>f</math> ''very convex'' if
+Call a real-valued [[function]] <math>f</math> ''very convex'' if
 <cmath>\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|</cmath>
-holds for all real numbers <math>x</math> and <math>y</math>. Prove that no very [[convex]] [[function]] exists.
+holds for all real numbers <math>x</math> and <math>y</math>. Prove that no very convex function exists.
 == Solution ==
+=== Solution 1 ===
 Let <math>y \ge x</math>, and substitute <math>a = x, 2b = y-x</math>. Then a function is very convex if <math>\frac{f(a) + f(a+2b)}{2} \ge f(a + b) + 2b</math>, or rearranging,
 <cmath>\left[\frac{f(a+2b)-f(a+b)}{b}\right]-\left[\frac{f(a+b)-f(a)}{b}\right] \ge 4</cmath>
 Let <math>g(a) = \frac{f(a+b) - f(a)}{b}</math>, which is the slope of the secant between <math>(a,f(a))(a+b,f(a+b))</math>. Let <math>b</math> be arbitrarily small; then it follows that <math>g(a+b) - g(a) > 4</math>, <math>g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4</math>. Summing these inequalities yields <math>g(a+kb)-g(a) > 4k</math>. As <math>k \rightarrow \infty</math> (but <math>b << \frac{1}{k}</math>, so <math>bk < \epsilon</math> is still arbitrarily small), we have <math>\lim_{k \rightarrow \infty} g(a+kb) - g(a) = g(a + \epsilon) - g(a) > \lim_{k \rightarrow \infty} 4k = \infty</math>. This implies that in the vicinity of any <math>a</math>, the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.
-== See also ==
-{{USAMO newbox|year=2000|before=First question|num-a=2}}
+=== Solution 2 ===
+Suppose, for the sake of contradiction, that there exists a very convex function <math>f.</math> Notice that <math>f(x)</math> is convex if and only if <math>f(x) + C</math> is convex, where <math>C</math> is a constant. Thus, we may set <math>f(0) = 0</math> for convenience.
+Suppose that <math>f(1) = A</math> and <math>f(-1) = B.</math> By the very convex condition,
+<cmath>\frac{f(0) + f\left(2^{-n}\right)}{2} \ge f\left(2^{-(n+1)}\right) + \frac{1}{2^n}.</cmath>
+A straightforward induction shows that:
+<cmath>f\left(2^{-n}\right) \le \frac{A - 2n}{2^n}</cmath>
+for all nonnegative integers <math>n.</math>
+Using a similar line of reasoning as above,
+<cmath>f\left(-2^{-n}\right) \le \frac{B - 2n}{2^n}.</cmath>
+Therefore, for every nonnegative integer <math>n,</math>
+<cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \le \frac{A+B-4n}{2^n}.</cmath>
+Now, we choose <math>n</math> large enough such that <math>n > \frac{A+B}{4} - 1.</math> This is always possible because <math>A</math> and <math>B</math> are fixed for any particular function <math>f.</math> It follows that:
+<cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) < \frac{1}{2^{n-2}}.</cmath>
+However, by the very convex condition,
+<cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.</cmath>
+This is a contradiction. It follows that no very convex function exists.
+== See Also ==
+{{USAMO newbox|year=2000|before=First Question|num-a=2}}
 [[Category:Olympiad Algebra Problems]]
+[[Category:Functional Equation Problems]]
+{{MAA Notice}}

2000 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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