It's well known that the continuous solutions to the Cauchy's functional equation:
$$ f ( x + y ) = f ( x ) + f ( y ) $$
are of the form $ f ( x ) = c x $ for some constant $ c $. However, I would like to know if the following generalization of the problem is true.
Suppose $ f $ is continuous, and there exist constants $ a < 0 < b $ such that
$$ a \le f ( x ) + f ( y ) – f ( x + y ) \le b $$
for every real $ x $ and $y$. Does this condition imply that there is a constant $ c $ such that
$$ a \le f ( x ) – c x \le b $$
for every real $ x $?
Best Answer
You can rewrite your inequlities in terms of $ g ( x ) = f ( x ) - \frac { a + b } 2 $ and $ \epsilon = \frac { b - a } 2 $, and ask for the following:
The answers to all these questions are positive. In fact this notion is well-known and has a name: stability. A good reference for stability of many famous functional equations is Hyers-Ulam-Rassias stability of functional equations in nonlinear analysis by S.M. Jung. I give the proof on page 21 of that book, with some minor changes.
The trick is to define $ A ( x ) = \lim _ { n \to \infty } 2 ^ { - n } g ( 2 ^ n x ) $. To show that the limit exists for every $ x $, first note that by \eqref{0}, $ | g ( 2 x ) - 2 g ( x ) | \le \epsilon $, or equivalently $ \big| \frac 1 2 g ( x ) - g \big( \frac x 2 \big) \big| \le \frac \epsilon 2 $ for every $ x $. It follows that $$ \left| 2 ^ { - n } g ( x ) - g \left( 2 ^ { - n } x \right) \right| = \left| \sum _ { i = 0 } ^ { n - 1 } \Big( 2 ^ { - n + i } g \big( 2 ^ { - i } x \big) - 2 ^ { - n + i + 1 } g \big( 2 ^ { - i - 1 } x \big) \Big) \right| \\ \le \sum _ { i = 0 } ^ { n - 1 } 2 ^ { - n + i + 1 } \big| 2 ^ { - 1 } g \big( 2 ^ { - i } x \big) - g \big( 2 ^ { - i - 1 } x \big) \big| \le \sum _ { i = 0 } ^ { n - 1 } 2 ^ { - n + i } \epsilon = ( 1 - 2 ^ { - n } ) \epsilon \text . \tag 2 \label 2 $$ Thus for $ m < n $ we get $$ | 2 ^ { - m } g ( 2 ^ m x ) - 2 ^ { - n } g ( 2 ^ n x ) | = 2 ^ { - m } | g ( 2 ^ m x ) - 2 ^ { m - n } g ( 2 ^ n x ) | \le 2 ^ { - m } ( 1 - 2 ^ { m - n } ) \epsilon = ( 2 ^ { - m } - 2 ^ { - n } ) \epsilon $$ which shows that $ \big( 2 ^ { - n } g ( 2 ^ n x ) \big) _ { n = 0 } ^ \infty $ is a Cauchy sequence and hence convergent. It follows from \eqref{0} that $ | g ( 2 ^ n x + 2 ^ n y ) - g ( 2 ^ n x ) - g ( 2 ^ n y ) | \le \epsilon $. Dividing by $ 2 ^ n $ and letting $ n \to \infty $ we see that $ A $ is an additive function. If we replace $ x $ by $ 2 ^ n x $ in \eqref{2} and take the limit, we have the inequality \eqref{1}.
Suppose that $ B $ is another additive function satisfying \eqref{1}. We can see that $$ | A ( x ) - B ( x ) | = \frac 1 n | A ( n x ) - B ( n x ) | \le \frac 1 n | A ( n x ) - g ( n x ) | + \frac 1 n | g ( n x ) - B ( n x ) | \le \frac { 2 \epsilon } n \text . $$ Hence $ B = A $, and $ A $ is the unique additive function satisfying the inequality \eqref{1}.
At last, we show that if $ g $ is continuous at any point $ x $, then $ A $ is continuous at $ 0 $, and since it's additive, continuous everywhere, which shows that it is linear. Since $ g $ is continuous at $ x $, there is a positive $ \delta $ such that if $ | y | < \delta $ then $ | g ( x + y ) - g ( x ) | < \epsilon $. We then have by \eqref{1} $$ | A ( y ) | = | A ( x + y ) - A ( x ) | \\ \le | A ( x + y ) - g ( x + y ) | + | g ( x + y ) - g ( x ) | + | g ( x ) - A ( x ) | < 3 \epsilon \text . $$ Since $ A $ is additive, we get that if $ | y | < \frac \delta n $ then $ | A ( y ) | < \frac { 3 \epsilon } n $, which shows that $ A $ is continuous at $ 0 $, and we're done.
Jung, Soon-Mo, Hyers-Ulam-Rassias stability of functional equations in nonlinear analysis, Springer Optimization and Its Applications 48. Berlin: Springer (ISBN 978-1-4419-9636-7/hbk; 978-1-4419-9637-4/ebook). xiii, 362 p. (2011). ZBL1221.39038.