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I need help solving the following functional equation.

Suppose $\,f : \mathbb{R} \to \mathbb{R}$ is differentiable and

$$f'\Big(\frac{x + y}{2}\Big) = \frac{f(x) - f(y)}{x -y}$$ holds for all $x \neq y$. Solve for $f$.

I can clearly see that any function of the form $f(x) = \alpha x$ solves this, but I don't know how to find any other solutions or show that this is the only one.

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  • $\begingroup$ Well $f(x) = \alpha x +1$ also works so I can tell you that there are more solutions to look for. I am not sure what they are yet though. $\endgroup$ – Josh R May 22 '16 at 8:15
  • $\begingroup$ $f(x) = \alpha x +\beta$ also works. $\endgroup$ – Claude Leibovici May 22 '16 at 8:35
  • $\begingroup$ @MathematicsStudent1122 Do you have a proof? $\endgroup$ – S.C.B. May 22 '16 at 8:43
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    $\begingroup$ $f(x)=x^2$ works, too. $\endgroup$ – Michael Hoppe May 22 '16 at 9:30
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    $\begingroup$ For context (and maybe future reference), I found this question in the book "Elementary Real Analysis" (classicalrealanalysis.info/com/Elementary-Real-Analysis.php), right after the section on the mean value theorem. $\endgroup$ – Anamaki May 22 '16 at 11:36
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The functional equation $$ f'\left(\frac{x + y}{2}\right) = \frac{f(y) - f(x)}{y - x}\quad\text{for all real $x$ and $y$,} $$ guarantees $f'$ is continuous: Fixing $x$ and letting $y = x + 2h \to x$, we have $$ \lim_{h \to 0} f'(x + h) = \lim_{h \to 0} \frac{f(x + 2h) - f(x)}{2h} = f'(x). $$ This technical fact is needed below.

For each real $z$ and for all $h \neq 0$, the functional equation says $$ 2h\, f'(z) = f(z + h) - f(z - h). $$ Fixing $z$ and differentiating with respect to $h$ (each side is differentiable in $h$ by hypothesis), $$ 2f'(z) = f'(z + h) + f'(z - h)\quad\text{for all real $z$ and $h$.} $$

In other words, $2f'(\frac{x + y}{2}) = f'(x) + f'(y)$ for all real $x$ and $y$. Since $f'$ is continuous, it follows that $f'$ is a linear polynomial, i.e., there exist real numbers $a$ and $b$ such that $$ f'(x) = 2ax + b\quad\text{for all real $x$,} $$ and consequently that $f$ is a quadratic polynomial.

This fact is well-known (and surely answered elsewhere on site), but in the spirit of being self-contained, here's an ad hoc sketch: If $f'(x)$ and $f'(y)$ are known for some $x < y$, then

  • $f'(z) = f'(x) + \dfrac{z - x}{y - x} \bigl(f'(y) - f'(x)\bigr)$ on $[x, y]$ by successive bisection and continuity;

  • $f'(2y - x) = f'(y + (y - x)) = 2f'(y) - f'(x)$ and $f'(y - 2x) = f'(y) - 2f'(x)$ are uniquely determined, so $f'$ is uniquely determined (hence linear) on the larger interval $[y - 2x, 2y - x]$. Inductively, $f'$ is linear on (arbitrary closed, bounded subintervals of) $(-\infty, \infty)$.


(Once the end result is known, more elegant arguments can doubtless be given, based on the fact that the functional equation is invariant if a quadratic polynomial is subtracted from $f$; interpolate $f$ at three points with a quadratic $ax^{2} + bx + c$, put $g(x) = f(x) - (ax^{2} + bx + c)$, and show $g \equiv 0$. I haven't carried out the details, however.)

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It's easy to show that any function $f(x)=ax^2+bx+c$ fulfills the equation. I remember that this result was discovered by Archimede.

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  • $\begingroup$ This also feels like a random guess, unfortunately. $\endgroup$ – Anamaki May 22 '16 at 11:30
  • $\begingroup$ Certainly not: en.m.wikipedia.org/wiki/The_Quadrature_of_the_Parabola $\endgroup$ – Michael Hoppe May 22 '16 at 11:45
  • $\begingroup$ I mean, it's not a "proof" that quadratics are the only solution to this. Just a bigger class of solutions which come from experience/looking and not from solving the question per se. $\endgroup$ – Anamaki May 22 '16 at 11:46
  • $\begingroup$ Well, it's easy to show that if $f$ is a polynomial, its degree is at most $2$. $\endgroup$ – Michael Hoppe May 22 '16 at 12:29
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You can maybe try restricting yourself to certain families. For example the following drafts an informal proof that, of the family of functions that can be expressed as Maclaurin series, only the polynomials up to 2nd order have this property. This family is quite big as it includes all polynomials, exp, sin, cos, log(1+x), ...

Write $f(x) = \sum_{i=0}^{\infty} \alpha_i x^i $. Replacing in your equation you should obtain $$\sum_{i=1}^{\infty} \frac {i \alpha_i}{2^{i-1}} (x+y)^{i-1} = \frac{ \sum_{i=0}^{\infty} \alpha_i x^i - \sum_{i=0}^{\infty} \alpha_i y^i}{x-y}.$$ Then multiply both sides by $(x-y) $: $$\sum_{i=1}^{\infty} \frac {i \alpha_i}{2^{i-1}} (x+y)^{i-1} (x-y) = \sum_{i=0}^{\infty} \alpha_i (x^i - y^i).$$ I'll now give the gist of the proof. We need to find values of $\alpha_i$ for which the above identity holds for all $x$ and $y$. This means that the coefficients of all monomials $x^jy^k$ must be the same on the lhs and hrs. We notice that on both sides $\alpha_i$ multiplies all and only monomials $x^jy^k$ such that $i=j+k$.

We have that $\alpha_0$ can be any real number since it doesn't appear on the lhs and it cancels out in the rhs. Similar conclusions can be drawn for $\alpha_1$ and $\alpha_2$ after expanding some products.

For $\alpha_i\; i\ge 3$, we notice that on the lhs this parameter multiples mixed monomials ($x^jy^k\; j, k > 0$) that are not present in the rhs. Hence $\alpha_i = 0\; i\ge 3$.

This proof is quite informal but I hope you got the main idea.

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