What about $f$ if $\frac{d\left(xf(x)-x\right)}{dx} \geq 0$

Question

I know that my function $f$ must satisfy the following condition for $x \geq 0:$ $$\frac{d\left(xf(x)-x\right)}{dx} \geq 0.$$ What can I say about $f?$ I am curious about its possible sign or variations with respect to $x$.

I have investigated Grönwall's inequality to find an upper bound for $f$ but it seems useless in my case.

Answer 1

HINT:

Note that

$$\begin{align} \frac{d}{dx}(xf(x)-x)&=xf'(x)+f(x)-1\\\\ &=x(f(x)-1)' +(f(x)-1) \end{align}$$

Now, let $u(x)=1-f(x)$ and $\beta(x)=-\frac1x$ and apply Gronwall's Inequality on the interval $[a,b]$ where $0<a<b$.

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Alternatively, simply integrate the given inequality over an interval for which it applies. What do you find directly about the bound on $f(x)$ in that interval without the need to appeal to Gronwall's Inequality?

Answer 2

From the mean value theorem it follows that if a function has positive derivative on an interval, then it is increasing there. Thus, with $x_2>x_1$ you can conclude that $$ x_2f(x_2)-x_2>x_1f(x_1)-x_1. $$ You can of course rearrange the terms if you want. If there is something more specific you want to say, then please update your question.