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If $f^{-1}(x)=kx-f(x)\forall x\in\mathbb{R}$ for a strictly increasing $f$ and $k$ a constant, then what can be said about $f$?

I think the answer is of the form $f(x)=x+c$, for some $c\in\mathbb{R}$. Any hints. Thanks beforehand

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    Since you define $f^{-1}$ and $f$ for all $x \in \mathbb{R}$, are you assuming that $f is also unbounded? – 2017-02-08
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    Does the definition mean $D_f=\mathbb{R}$ – 2017-02-08
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    @MyGlasses yes the domain the set of real numbers. – 2017-02-09
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    At least some solutions can be found here http://math.stackexchange.com/questions/2122305/variant-of-cauchy-functional-equation/2132515#2132515 – 2017-02-15

1 Answers 1

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(1) $f$ is convex iff $f^{-1}$ is concave.

${\bf Proof.}$ Let $x_1, x_2\in D_f$ and $\lambda\in\mathbb{R}$ so with $g=f^{-1}$ \begin{eqnarray} g(\lambda x_1+(1-\lambda)x_2)&=&k(\lambda x_1+(1-\lambda)x_2)-f(\lambda x_1+(1-\lambda)x_2)\\ &\geq& k\lambda x_1+k(1-\lambda)x_2-\lambda f(x_1)-(1-\lambda)f(x_2)\\ &=& \lambda g(x_1)+(1-\lambda)g(x_2) \end{eqnarray}

(2) If $k<0$ then $g=f^{-1}$ is decreasing.

${\bf Proof.}$ Let $x_1, x_2\in D_f$ and $x_1kx_2$ hence $g(x_1)>g(x_2)$.

(3) If $k\neq 2$ then $f(x)\neq x+c$ for a $c\in\mathbb{R}$.

${\bf Proof.}$ If $f(x)=x+c$ then $f^{-1}(x)=x-c$ and $x-c=kx-x-c$ therefore $k=2$.

(4) If $f(x)$ be continuous, then $$\int f^{-1}(x)dx=\frac12kx^2-\int f(x)dx+C$$ integration by parts shows $\int f^{-1}(x)dx=xf(x)-\int f(x)dx$ thus $$f(x)=\frac12kx+\frac{C}{x}$$

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    how can we prove that? – 2017-02-08
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    how can you say that $f$ is differentiable? – 2017-02-08
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    still, how can you prove that? – 2017-02-08
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    If $k = 0$, $x = f^{-1}(f(x)) = -f(f(x))$. Can that have a real solution? I can't think of any way it could, especially given the stipulation that $f$ is strictly increasing. Does that mean $k \not=0$? – 2017-02-08
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    thanks for your reply, by the way is the converse of $(3)$ in your answer valid? – 2017-02-09
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    I don't understand how you came to conclusion (4). Here a conflicting result: For $k^2 \neq 4 $ we have : $f(x)= (\frac{k}{2}\pm \sqrt{\frac{k^2}{4}-1} )\cdot x $ $\implies f^{-1}(x)= \frac{x}{( \frac{k}{2} \pm \sqrt{\frac{k^2}{4}-1} )}$ $\implies f^{-1}(x)+f(x)= \frac{x}{( \frac{k}{2} \pm \sqrt{\frac{k^2}{4}-1} )}+(\frac{k}{2}\pm \sqrt{\frac{k^2}{4}-1} )\cdot x$ $\implies f^{-1}(x)+f(x)= \frac{( \frac{k}{2} \pm \sqrt{\frac{k^2}{4}-1} )}{( \frac{k}{2} \pm \sqrt{\frac{k^2}{4}-1} )}\cdot kx$ – 2017-02-15
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    What I meant was: $(\frac{k}{2}\pm \sqrt{\frac{k^2}{4}-1} ) \neq \frac{k}{2}$ – 2017-02-15
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    @RutgerMoody Maybe you right, I think (4) is wrong. – 2017-02-15