# Cauchy's functional equation with bounds: $a \le f ( x ) + f ( y ) - f ( x + y ) \le b$

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 $$x$$?

• At least case with $f(x)+f(y)-f(x+y)=d$ for $d$ constant should be easy, as it turns into Cauchy's equation by $g(x)=f(x)-d$ and solves to $f(x)=cx+d$. – Sil Apr 15 '20 at 12:44

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:

If for some nonnegative real $$\epsilon$$ $$| g ( x + y ) - g ( x ) - g ( y ) | \le \epsilon \tag 0 \label 0$$ for every real $$x$$ and $$y$$, is there an additive function $$A$$ such that $$| g ( x ) - A ( x ) | \le \epsilon \tag 1 \label 1$$ for all $$x$$? Is such $$A$$ unique? Does continuity of $$g$$ imply linearity of $$A$$?

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.