find all $f:\mathbb{R}\to\mathbb{R}$ so that $|f(x+y)-f(x-y) - y|\leq y^2$ for all $x,y\in\mathbb{R}$

Question

Find all $f:\mathbb{R}\to\mathbb{R}$ so that $|f(x+y)-f(x-y) - y|\leq y^2$ for all $x,y\in\mathbb{R}$.

A standard approach when solving functional equations or functional inequalities is to guess which function might work, and then show that the guessed function is unique. One can show uniqueness by showing that if $g$ is the guessed function and $f$ is any function satisfying the given conditions, then $f-g$ is identically zero. The function $f(x)=x/2$ clearly works. It's definitely the only linear function with zero constant term that works. Note that we may assume $f(0) = 0$ because if we let $h(x)=f(x)+C$ for an arbitrary constant $C$, then $h$ satisfies the same inequality as $f$. In general, any function satisfying the given inequality is the sum of a constant and a function $f$ satisfying the inequality and $f(0) = 0$. If $f(x)=cx$ for all x then the inequality reads $|c(x+y) - c(x-y) - y|\leq y^2\,\forall x,y\Rightarrow \,\forall y, |y(2c-1)|\leq y^2.$ But as $y\to 0,|\dfrac{y^2}{y}|\to 0$ so eventually $y^2$ will be less than $|y(2c-1)|$ for sufficiently small y, and hence $2c-1$ must be equal to $0$. Thus we've shown $c=1/2$ if $f(x)=cx$ for all $x$. Let $g(x)=f(x)-x/2.$ It could be useful to show $g(x)=0$ for all x. We clearly have $|g(x+y)-g(x-y)| = |f(x+y)-(x+y)/2 - f(x-y) +(x-y)/2| = |f(x+y)-f(x-y) - y|\leq y^2$ for all $x,y\in\mathbb{R}$. Plugging in $x=y$ gives that $|g(2y)-g(0)| \leq y^2$ for all $y$ and by our assumption, $g(0)$ can be assumed to be zero, so $|g(2y)|\leq y^2$ for all $y$. But I'm not sure how to show that $g$ must be identically zero, assuming $f(0) = 0$.

Answer 1

You have made great progress.

Here are two further ideas.

• Construct the supposedly-identically-zero function. Then try proving it is indeed identically zero. We love zero.
• Estimate the change of function over an interval by splitting the interval into smaller and smaller intervals.

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Let $g(x)=f(x)-(x/2+f(0))$. Then $g(0)=0$. $$\begin{aligned}&g(x+y)-g(x-y)\\ =&f(x+y)-((x+y)/2+f(0))-(f(x-y)-((x-y)/2+f(0)))\\ =&f(x+y)-f(x-y) - y \end{aligned}$$

Hence, we have $$|g(x+y)-g(x-y)|\le y^2$$ for all $x, y\in\Bbb R$. Replacing $(x,y)$ by $((x+y)/2, (x-y)/2)$, we have $$ |g(x)-g(y)|\le((x-y)/2)^2$$

For any $x\in\Bbb R$, $$\begin{aligned}&|g(x)-g(0)|\\ =&\left|\sum_{i=1}^n |g(ix/n)-g((i-1)x/n)\right|\\ \le&\sum_{i=1}^n |g(ix/n)-g((i-1)x/n)|\\ \le&\sum_{i=1}^n ((ix/n-(i-1)x/n)/2)^2\\ =&\sum_{i=1}^n (\frac x{2n})^2\\ =&\frac {x^2}{4n} \end{aligned}$$ Since $n$ can be arbitrarily large, the inequality above implies $|g(x)-g(0)|$ must be smaller than any positive number. That means it must be $0$. Hence $g(x)=0$ for all $x$. $f(x)=x/2+f(0)$ for all $x$.

_{This answer is prepared especially for students who have not learned calculus. For people who have, the given condition implies immediately the derivative of f(x) is $\frac12$ everywhere.}

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Here is a minor variation as an easy exercise.

Find all $f:\mathbb{R}\to\mathbb{R}$ so that $|f(x+2y)-f(x-3y) - y|\leq |y|^{\frac32}$ for all $x,y\in\mathbb{R}$.

Answer 2

Dividing by $\vert y\vert$ (for $y\neq0$), we get : $$\left|\frac{f(x+y)-f(x)}{y}+\frac{f(x)-f(x-y)}{y}-1\right|\leqslant\vert y\vert$$ Assuming that $f$ is differentiable, we get, as $y\rightarrow 0$, that for all $x$ : $$f'(x)=\dfrac12$$ Hence, there exists $C\in\mathbb{R}$ such that $f(x)=\dfrac x2+C$ for all $x$. And, as noted in the OP, all those functions are différentiable and verify the requested condition.

It remains to show that diffentiability is mandatory ...

Answer 3

It's really the definition of derivative in disguise... if you replace $x$ by $x + y$ then $y$ by ${y \over 2}$, then your condition is that $$\bigg|{f(x + y) - f(x) \over y} - {1 \over 2}\bigg| \leq {y \over 4}$$ So as a result $${1 \over 2} - {y \over 4} \leq {f(x + y) - f(x) \over y} \leq {1 \over 2} + {y \over 4}$$

By the squeeze test you can take limits as $y$ goes to zero and get that $f'(x) = {1 \over 2}$ for all $x$ and conclude that $f(x) = {x \over 2} + C$ for some constant $C$. It can be easily verified that each such function works, so those are the functions that satisfy the equation.

Answer 4

The answer is $f(t) = \frac{t}{2}+C$ where $C$ is any constant.

Proof.

Any two numbers $a$ and $b$ can be expressed as $b=x+y,a=x-y$. Therefore, the condition is equivalent to

$$|f(b)-f(a)-(b-a)/2|\le (b-a)^2/4.$$

Let $g(t)=f(t)-t/2$. Then we can rewrite the inequality as

$$|g(b)-g(a)|\le (b-a)^2/4.$$

Fixing $a$ and letting $b\to a$, this inequality implies $g$ is differentiable at $a$ and its derivative is zero. Therefore $g$ is a constant function. That is $f(t) = t/2+C$. Clearly, with this choice (any $C$!) the original inequality holds, therefore we have found all solutions.