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I'm stuck with a problem here, because i could not justify my way around a situation.

THE PROBLEM

Let $f:R \to R$ be a function satisfying:

$|f(x+y) -f(x-y) -y| \le y^2$ for all $x,y \in R$. Then show that

$f(x) = \frac{x}{2} + c$ for some constant c .

MY DOUBT

I started by putting $x=y$ which gives,

$|f(2x) - f(0) - x| \le x^2$

However, this eqn raised a doubt. Suppose $f(2x) = x^2 + c + x$ then the inequality willbeconverted into an equality,so the conditions are satisfied. But this is in direct contradiction with what we are asked to prove. Can someone please point out a mistake, because i coudnt find any when i tried?

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    Did you notice my inclusion in previous answer of an extra section titled "Added" in which I give what I hope is a proof of $f(x)=x/2+c$? Let me know if any questions about that... – 2017-01-29
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    I just saw that now. Thanks a lot for your help :)! – 2017-01-29

3 Answers 3

2

By letting $x=y$ you're only looking at a subset of the full collection of pairs $(x,y)$ for which the inequality is assumed. So that is criticism 1.

Also, there is no "right" to replace the $\le$ by $=.$ Or in other words, you have really just defined a function of $x$ alone, and noted that it satisfies the result of substituting $x=y$ in the original inequality.

What you really need to find is a function $f$ which satisfies the given inequality for all possible pairs $(x,y).$

Added: a proof.

The assumed inequality is $$|f(x+y)-f(x-y)-y|\le y^2 \tag{1}$$ for all real $x,y.$ Let $u=x+y,\ v=x-y$ so that $y=(u-v)/2.$ Now assume $u-v>0$ and divide $(1)$ by $u-v,$ to get $$|(f(u)-f(v))/(u-v)-1/2|\le(u-v)/4. \tag{2}$$ If we now let $u \to v^+$ in $(2)$ we see that the right hand derivative of $f$ at the arbitrary real number $v$ exists and is $1/2.$ On the other hand if we let $v \to u^-$ in $(2)$ we see that the left hand derivative of $f$ at the arbitrary real number $u$ exists and is also $1/2.$

Since here each of $u,v$ are arbitrary, and both one-sided derivatives agree, we have that the (two sided) derivative of $f$ exists and is $1/2$ everywhere. Thus $f'(x)=1/2$ so that $f(x)=(1/2)x+c$ for some real $c$ as required.

2

Let $f(t)=\frac{t}2+g(t)$. Then the functional inequality can be rewritten as:

\begin{align} &\left|\frac{x+y}{2}+g(x+y)-\left(\frac{x-y}{2}+g(x-y)\right)-y\right|\leq y^2\\ \implies&|g(x+y)-g(x-y)|\leq y^2. \end{align}

Let $x-y=a$ and $2y = b$. Then, whenever $y\neq0$,

$$|g(a+b)-g(a)|\leq\frac{b^2}{4}\implies\left|\frac{g(a+b)-g(a)}{b}\right|\leq \frac{|b|}4.$$

Using this, what can you deduce about $g$ as $b\to 0$?

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    I think this is like what I have "added" to my earlier answer. – 2017-01-29
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    Yes, exactly! I actually found this problem quite interesting. – 2017-01-29
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    I noticed that perhaps your answer may have preceded mine in when it actually got put up. I also liked the problem, and will upvote your answer... – 2017-01-29
1

Your function satisfied your second equation, and the first one when $x=y$. But it doesn't satisfy the fist equation for all $x,y\in\mathbb R$, which is a stronger condition than the one with $x=y$