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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<cmath>f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.</cmath> | <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. | This is a contradiction. It follows that no very convex function exists. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Define the symmetric average | ||
+ | <cmath> | ||
+ | F(x, y) = \frac{f(x) + f(y)}{2}. | ||
+ | </cmath> | ||
+ | Then the inequality becomes | ||
+ | <cmath> | ||
+ | F(x, y) \ge f\left( \frac{x + y}{2} \right) + |x - y|. | ||
+ | </cmath> | ||
+ | |||
+ | Because <math>|x - y|</math> is not differentiable along the line <math>x = y</math>, the surface defined by <math>F(x, y)</math> must have a permanent kink along the diagonal. A kink is a sharp crease where the surface fails to have a well-defined tangent plane. | ||
+ | |||
+ | If f were a true function on <math>\mathbb{R}</math> satisfying the inequality at all scales, then this kink would have to persist no matter how closely x and y are chosen. That means the graph of F never becomes smooth near the diagonal. | ||
+ | |||
+ | But every differentiable real-valued function f produces a smooth symmetric average F. Therefore, the presence of a permanent kink contradicts the assumption that f is differentiable at any scale. The kink cannot be removed by local smoothing. | ||
+ | |||
+ | Hence, the geometric shape of the inequality forces a kink that no function f can produce, so no function can be very convex. | ||
+ | |||
+ | ~Lopkiloinm | ||
== See Also == | == See Also == |
Latest revision as of 03:48, 30 June 2025
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.
Solution 3
Define the symmetric average
Then the inequality becomes
Because is not differentiable along the line
, the surface defined by
must have a permanent kink along the diagonal. A kink is a sharp crease where the surface fails to have a well-defined tangent plane.
If f were a true function on satisfying the inequality at all scales, then this kink would have to persist no matter how closely x and y are chosen. That means the graph of F never becomes smooth near the diagonal.
But every differentiable real-valued function f produces a smooth symmetric average F. Therefore, the presence of a permanent kink contradicts the assumption that f is differentiable at any scale. The kink cannot be removed by local smoothing.
Hence, the geometric shape of the inequality forces a kink that no function f can produce, so no function can be very convex.
~Lopkiloinm
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.