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

  • $\begingroup$ 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

(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 $$f(x)=\frac12kx+\frac{C}{x}$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.