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2000 USAMO Problems/Problem 1: Difference between revisions

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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,
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>
<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.
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.
 
 
== 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,
<math>\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}</math><math>
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.$ 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 there exists no very convex function.


== See Also ==
== See Also ==

Revision as of 23:55, 4 April 2016

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 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}.<cmath> A straightforward induction shows that: </cmath>f\left(2^{-n}\right) \le \frac{A - 2n}{2^n}$$for all nonnegative integers$n.$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$n,$<cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \le \frac{A+B-4n}{2^n}.</cmath> 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 there exists no very convex function.

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

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