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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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  • $\begingroup$ Since you define $f^{-1}$ and $f$ for all $x \in \mathbb{R}$, are you assuming that $f is also unbounded? $\endgroup$ – Badam Baplan Feb 8 '17 at 7:47
  • $\begingroup$ Does the definition mean $D_f=\mathbb{R}$ $\endgroup$ – Nosrati Feb 8 '17 at 7:49
  • $\begingroup$ @MyGlasses yes the domain the set of real numbers. $\endgroup$ – vidyarthi Feb 9 '17 at 7:24
  • $\begingroup$ At least some solutions can be found here math.stackexchange.com/questions/2122305/… $\endgroup$ – Rutger Moody Feb 15 '17 at 15:34
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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 $x_1<x_2$ then $f(x_1)<f(x_2)$ and $kx_1>kx_2$ 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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  • $\begingroup$ how can we prove that? $\endgroup$ – vidyarthi Feb 8 '17 at 6:31
  • $\begingroup$ how can you say that $f$ is differentiable? $\endgroup$ – vidyarthi Feb 8 '17 at 6:32
  • $\begingroup$ still, how can you prove that? $\endgroup$ – vidyarthi Feb 8 '17 at 6:35
  • $\begingroup$ 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 $k \not=0$? $\endgroup$ – Badam Baplan Feb 8 '17 at 7:06
  • $\begingroup$ thanks for your reply, by the way is the converse of $(3)$ in your answer valid? $\endgroup$ – vidyarthi Feb 9 '17 at 7:19

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