Functional Equations: Cauchy's Method

In one sentence

Cauchy’s method is a ladder. The equation $f (x + y) = f (x) + f (y)$ first tells you $f$ on the positive integers; with $f (0)$ and $f (- x)$ you climb to the integers; with “ $q$ copies of $\frac{p}{q}$ add up to $p$ ” you reach the rationals; finally the monotonicity or continuity given in the problem lets the rationals squeeze the irrationals.

Solve f(x + y) = f(x) + f(y) for increasing f

Step 1 / 6

ℕ: add one step at a time
Write $c = f (1)$ . Put $y = 1$ : $f (n + 1) = f (n) + c$ . Also $x = y = 0$ gives $f (0) = 0$ , so $f (n) = n c$ .

$f (n) = n c$
ℤ: put y = −x
ℚ: q copies of p/q add up to p
ℝ: the equation cannot reach irrationals
Squeeze with monotonicity
Finish by contradiction

Predict first

$f (x + y) = f (x) + f (y)$ for all real numbers, and $f (1) = 3$ . What is $f (\frac{2}{3})$ ?

Choose an answer first; then you can see the result.

Stop 1 · Exponential type: $f (x + y) = f (x) f (y)$

“Add” has become “multiply”. First look at the sign: $f (x) = f (\frac{x}{2})^{2} \geq 0$ . If $f (x_{0}) = 0$ somewhere, then $f (x) = f (x_{0}) f (x - x_{0}) = 0$ everywhere. Otherwise $f > 0$ , and you can take logarithms:

g (x) = ln f (x) ⟹ g (x + y) = g (x) + g (y) .

Back to Cauchy’s equation! If $f$ is monotonic or continuous, $g (x) = x ln a$ , that is, $f (x) = a^{x}$ with $a = f (1) > 0$ .

Stop 2 · Logarithmic type: $f (x y) = f (x) + f (y)$ ( $x, y > 0$ )

Substitute $x = e^{u}$ , $y = e^{v}$ and let $g (u) = f (e^{u})$ . You get $g (u + v) = g (u) + g (v)$ . Same ladder; the conclusion is $f (x) = k ln x$ , that is, a multiple of $lo g_{a} x$ .

Stop 3 · Midpoint type: $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$

Turning the midpoint equation into Cauchy’s equation

Step 1 / 4

Shift first
Let $g (x) = f (x) - f (0)$ . Then $g$ satisfies the same equation, and $g (0) = 0$ .

$g (x) = f (x) - f (0)$
Put y = 0
It becomes Cauchy’s equation
Conclusion

Remove the extra condition

Break it

Counterexample lab: remove one condition

Conclusion: the only solutions of $f (x + y) = f (x) + f (y)$ are $f (x) = c x$

With monotonicity, the irrational number $π$ is tightly squeezed by rationals on both sides. Raise the denominator bound $q$ and watch the box shrink.

Practice

Exercise 1Basic

$f (x + y) = f (x) + f (y)$ for all real numbers, and $f (1) = 2$ . Find $f (- \frac{3}{4})$ .

Exercise 2Basic

$f (x + y) = f (x) + f (y)$ for all real numbers, and $f (3) = 5$ . Find $f (\frac{1}{2})$ .

Exercise 3Advanced

$f (x + y) = f (x) f (y)$ for all real numbers, and $f (1) = 4$ . Find $f (\frac{3}{2})$ .

Exercise 4Challenge

A continuous function $f$ satisfies $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$ , with $f (0) = 1$ and $f (2) = 5$ . Find $f (7)$ .

In one sentence

Solve f(x + y) = f(x) + f(y) for increasing f

Stop 1 · Exponential type: $f (x + y) = f (x) f (y)$

Stop 2 · Logarithmic type: $f (x y) = f (x) + f (y)$ ( $x, y > 0$ )

Stop 3 · Midpoint type: $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$

Turning the midpoint equation into Cauchy’s equation

Remove the extra condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Proving it for the rationals, then claiming it for all reals

Functional Equations: Cauchy's Method

In one sentence

Stop 1 · Exponential type: f(x+y)=f(x)f(y)

Stop 2 · Logarithmic type: f(xy)=f(x)+f(y) (x,y>0)

Stop 3 · Midpoint type: f(2x+y​)=2f(x)+f(y)​

Remove the extra condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Stop 1 · Exponential type: $f (x + y) = f (x) f (y)$

Stop 2 · Logarithmic type: $f (x y) = f (x) + f (y)$ ( $x, y > 0$ )

Stop 3 · Midpoint type: $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$