Difference between revisions of "2000 USAMO Problems/Problem 1"
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== Problem == | == 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> | <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 | + | holds for all real numbers <math>x</math> and <math>y</math>. Prove that no very convex function exists. |
− | == Solution 1 == | + | == 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, | 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, | ||
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− | == Solution 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, 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. | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:45, 9 July 2017
Problem
Call a real-valued function very convex if
holds for all real numbers and . Prove that no very convex function exists.
Solution
Solution 1
Let , and substitute . Then a function is very convex if , or rearranging,
Let , which is the slope of the secant between . Let be arbitrarily small; then it follows that , . Summing these inequalities yields . As (but , so is still arbitrarily small), we have . This implies that in the vicinity of any , 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 Notice that is convex if and only if is convex, where is a constant. Thus, we may set for convenience.
Suppose that and By the very convex condition, A straightforward induction shows that: for all nonnegative integers Using a similar line of reasoning as above, Therefore, for every nonnegative integer Now, we choose large enough such that This is always possible because and are fixed for any particular function It follows that: However, by the very convex condition, This is a contradiction. It follows that no very convex function exists.
See Also
2000 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.