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Let $f(x,y)$ be multi variable function that is defined when $\frac{x}{2}<y <x $ , and I want to know if there is such a function that satisfies :

1 ) $f(x-1,y) > f(x,y)$

2 ) $f(x-1,y-\ln x) > f(x,y)(1-\frac{1}{x})$

3 ) $|f(x,y) - \ln x| < \frac{1}{\sqrt{x}}$ when $|y- x| < 3\ln x$, this condition can be loosened to say $\frac{1}{x^{\frac{1}{3}}}$ or more (but for a fixed constant power).

I think such function do not exist, I just don't have a proof.

Please give a proof that such function does not exist or an example for such a function

Edit : I am interested as $x \to \infty$ how the function behave, also i am assuming that $f(x-1,y),f(x-1,y-\ln x),f(x,y)$ are all defined.

For example the function $f(x,y) = \ln (x) +1-\frac{x}{y}$ does satisfy the conditions $1,3$ but does not satisfy $2$.

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  • $\begingroup$ If the domain of $f$ is the region characterized by $x/2 < y < x$, then it's possible that both $(1)$ and $(2)$ are meaningless statements. For example, if $(x, y) = (1, 3/4)$, then $(x,y)$ is in the domain, but $(x-1, y) = (0, 3/4)$ is not in the domain, so the expression $f(x-1, y)$ is undefined. Maybe we should clarify what's meant by the expression $f(x-1, y)$ in these situations? $\endgroup$ – Justin Le May 2 at 23:26
  • $\begingroup$ @JustinLe assuming that $f(x,y),f(x-1,y),f(x-1,y-\ln x)$ are defined, i really don't care about the domain as long as $y$ is less than $x$ and the 3 conditions are satisfied. $\endgroup$ – Ahmad May 2 at 23:53
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    $\begingroup$ What motivation has this problem? $\endgroup$ – Vítězslav Štembera May 8 at 6:54
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HINT

$\color{brown}{\textbf{Domains and unknowns.}}$

Taking in account condition $$\frac x2 < y < x$$ expressions $f(x,y)$ and $f(x-1)$ exists in the same time if $$\frac x2 < y < x-1,\quad x \in [2,\infty).\tag1$$

Since $$\dfrac x2-n\ln x = \dfrac t2-n\ln t + \left(\dfrac12 -\dfrac nt\right)(x-t) +\dfrac n{2t^2}(x-t)^2 - \dfrac n{3t^3}(x-t)^3+\dots,$$ then $y-n\ln x > g_n(x,t),$ where $$g_n(x,t) = \dfrac t2-n\ln t + \left(\dfrac12 -\dfrac nt\right)(x-t) +\dfrac {3n}{2t}\dfrac{(x-t)^2}{2x+t}.\tag2$$ In particular, $y-\ln x > g_1(x,2),$ or $$y-\ln x > \ln\frac e2 + \frac38\,\frac{(x-2)^2}{x+1},\quad\text{where}\quad x>2,\tag{3a}$$ with the angular coefficient from zero to ${\large\frac38}$ (see also Wolfram Alpha plot).

Plot x/2-ln x >g2(x,2)

The slope $k=g'_2(t) = {\large \frac38 \frac{(t-2)(t+4)}{(t+1)^2}}$ of the tangent line $AB,$ where $A=(0,2)$ and $B=(t,g(t)),$ can be obtained from the condition $g(t) = k(t-2),\phantom{\Big|}$ or $$\ln\frac e2 +\dfrac38 \left(\dfrac{t-2}{t+1}\right)^2(t+1-(t+4)) = 0,$$ with the solution $$t = \frac{q+2}{1-q},\quad k = \frac38 \frac{(t-2)(t+4)}{(t+1)^2},\quad\text{where}\quad q=\frac{2\sqrt{2\ln{\Large\frac e2}\phantom{\bigg|}}}3, \quad \dbinom tk \approx \dbinom{5.279}{0.289\,412}.$$ Then $$\dfrac{y-\ln x}{x-2}\in \begin{cases}\left(\dfrac27,\infty\right),\quad\text{if}\quad x\in(2,\infty)\\ \left(\dfrac27,\dfrac12\right),\quad\text{if}\quad x\in(3,\infty). \end{cases}\tag{3b}$$

Plot x/2-ln x, slopes

Domain of the third condition $$y>x-3\ln x$$ is actual if $x> 3\ln x,$ or $x>u,$ where $$u = e^{-\large W_{\large-1}\left({\large-\frac13}\right)}\approx 4.536\,403\,655,\tag4$$ where $W_{-1}(x)$ is the analytic continuation of Lambert $W$ function (see also Wolfram Alpha solution and calculations).

Then $x-3\ln x > 2g_{3/2}(x,u),$ or $$x-3\ln x>\frac{x-u}{2u}\,\frac{(4u-3)(x-u)+6u(u-3)}{2x+u},\quad\text{if}\quad x\in(u,\infty),\tag{5a}$$ with the angular coefficient changing ${\large\frac{u-3}{u}}\approx0.339$ to ${\large\frac{4u-3}{4u}}\approx0.835$ (see also Wolfram Alpha plot).

x-3ln x

Then $$\dfrac{x-3\ln x}{x-u}\in \left(\dfrac{1}{3},1\right), \quad\text{if}\quad x\in(u,\infty).\tag{5b}$$

$\color{brown}{\textbf{Mltiplicative model.}}$

Let us try to search function $f$ in the multiplicative form of

$$f(x,y) = X(x)Y(y).\tag6$$

Taking in account $(1)-(5),$ let us consider the system of conditions in the form of $$\begin{cases} X(x-1)\,Y(y) > X(x)\,Y(y),\quad\text{if}\quad x>2\\[4pt] X(x-1)\,Y\left(y-\ln x\right)> \dfrac{x-1}x\,X(x)\,Y(y),\quad\text{if}\quad x>2\\[4pt] |X(x) Y(x-3\ln x) - \ln x| < \dfrac1{\sqrt x},\quad\text{if}\quad x>u\\[4pt] |X(x) Y(x-1) -\ln x| < \dfrac1{\sqrt x},\quad\text{if}\quad x>u\\ y\in\left(\dfrac x2,x-1\right)\\ y-\ln x \in\left(\dfrac{2(x-2)}7, \dfrac{x-2}2\right)\\ x-3\ln x \in \left(\dfrac{x-u}3, \dfrac{x-u}2\right), \end{cases}\tag7$$ where $u$ is given by $(4).$

Assume $X(x)$ and $Y(y)$ monotonic positive functions.

Easily to see that the first condition is satisfied if the function $X(x)$ decreases in the interval $(2,\infty).$

Taking in account $(1),$ the third condition in the form $(7.4)$ means that the production $X(x)Y(y)$ infinitely increases when $x\to\infty.$ Thus, function $Y(y)$ increases in the interval $(1,\infty).$

Taking in account increasing factor $\frac{x-1}x$ in the $\textbf{second condition},$ function $X(x)$ should contain decreasing factor ${\large\frac1x}.$ Taking in account $(7.4),$ it should contain logarithmic factor. It is possible, because the production $x^{-1}\ln^p(x+1)$ decreases in the interval $x\in(1,\infty)$ if $p\in(0,1].$

Taking in account $(3a),$ looks that the functions $$X(x) = \left(\ln(x+1)\right)^p x^{-q},\quad Y(y)= C\, \left(\ln(x+1)\right)^r(2y)^{-s}\tag8$$ can be solution for certain positive values of $p,q,r,s,$

Also, can be useful function $\dfrac{\ln^p\left(x+q\right)}{x},$ which decreases if $p\in[0,2],\,q\ge \dfrac{223}{168}\, \left(q=\dfrac43\right).$

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Version of 13.05.19

$\color{brown}{\textbf{Domains and unknowns.}}$

Taking in account condition $$\frac x2 < y < x$$ expressions $f(x,y)$ and $f(x-1)$ exists in the same time if $$\frac x2 < y < x-1,\quad x \in [2,\infty).\tag1$$

Taylor series for logarithmic function at $x=t$ is $$\ln x = \ln t + \dfrac{x-t}t - \dfrac{(x-t)^2}{2t^2} + \dfrac{(x-t)^3}{3t^3}+\dots,$$ so $$\ln x\in[\lambda(x,t),\Lambda(x,t)],\tag{L1}$$ where $$\lambda(x,t) =\ln t +\dfrac1t\dfrac{x-t}{1+{\large\frac{x-t}{2t}}} = \ln t + 2\dfrac{x-t}{x+t},$$ $$\Lambda(x,t) = \ln t + \dfrac{x-t}t - \dfrac1{2t^2}\dfrac{(x-t)^2}{\large1+\frac23\frac{x-t}t} = \ln t + \dfrac1t -\dfrac3{2t}\dfrac{(x-t)^2}{2x+t}$$ $$= \ln t +\dfrac{x-t}{2t}\dfrac{2(2x+t)-3(x-t)}{2x+t} = \ln t +\dfrac{(x-t)(x+5t)}{2t(2x+t)},$$

$$\lambda(x,t) = \ln t + 2\dfrac{x-t}{x+t}, \quad \Lambda(x,t) = \ln t +\dfrac{(x-t)(x+5t)}{2t(2x+t)}.\tag{L2}$$ $$\lambda(x,2) = \ln 2 + 2\dfrac{x-2}{x+2}, \quad \Lambda(x,t) = \ln 2 +\dfrac{(x-2)(x+10)}{8(x+1)}.\tag{L3}$$ Then $$\dfrac x2 - \ln x \in\left(\ln\dfrac e2 +\dfrac38\dfrac{(x-2)^2}{x+1},\, \ln\dfrac e2 +\dfrac12\dfrac{(x-2)^2}{x+2}\right),\quad x\in(2,\infty)\tag{2}$$ (see also Wolfram Alpha plot),

x/2-ln x

$$x-1 - \ln x \in\left(-\ln2e +\dfrac18\dfrac{7x^2+20}{x+1},\, -\ln2e +\dfrac{x^2+4}{x+2}\right),\quad x\in(2,\infty)\tag{3}$$ (see also Wolfram Alpha plot).

x-1-lnx

Domain of the third condition $$y>x-3\ln x$$ is actual if $x> 3\ln x,$ or $x>u,$ where $$u = e^{-\large W_{\large-1}\left({\large-\frac13}\right)}\approx 4.536\,403\,655,\tag4$$ where $W_{-1}(x)$ is the analytic continuation of Lambert $W$ function (see also Wolfram Alpha solution and calculations).

Taking in account $(\mathrm L2),$ one can write $$x-3\Lambda(x,u) = x - 3\left(\ln u +\dfrac{(x-u)(x+5u)}{2u(2x+u)}\right) = x - u - 3\dfrac{(x-u)(x+5u)}{2u(2x+u)} = \dfrac{(x-u)(x-7u)}{2u(2x+u)},$$

$$x-3\lambda(x,u) = x - 3\left(\ln u + 2\dfrac{x-u}{x+u}\right) = x-u-6\dfrac{x-u}{x+u} = \dfrac{(x-u)(x+u-6)}{x+u}$$

Then $x-3\ln x > 2g_{3/2}(x,u),$ or $$x-3\ln x \in\left(\dfrac{(x-u)(x-7u)}{2u(2x+u)},\dfrac{(x-u)(x+u-6)}{x+u}\right),\quad\text{if}\quad x\in(u,\infty),\tag{5}$$ (see also Wolfram Alpha plot).

x-3ln x

$\color{brown}{\textbf{Mltiplicative model.}}$

Let us try to search function $f$ in the multiplicative form of

$$f(x,y) = X(x)Y(y).\tag6$$

Taking in account $(1)-(5),$ let us consider the system of conditions in the form of $$\begin{cases} X(x-1)\,Y(y) > X(x)\,Y(y),\quad\text{if}\quad x>2\\[4pt] X(x-1)\,Y\left(\dfrac x2-\ln x\right) > \dfrac{x-1}x\,X(x)\,Y(\dfrac x2), \quad\text{if}\quad x>2\\[4pt] X(x-1)\,Y\left(x-1-\ln x\right)> \dfrac{x-1}x\,X(x)\,Y(x-1), \quad\text{if}\quad x>2\\[4pt] |X(x) Y(x-3\ln x) - \ln x| < \dfrac1{\sqrt x},\quad\text{if}\quad x>u\\[4pt] |X(x) Y(x-1) -\ln x| < \dfrac1{\sqrt x},\quad\text{if}\quad x>u\\ y\in\left(\dfrac x2,x-1\right)\\ \dfrac x2 - \ln x \in\left(\ln\dfrac e2 +\dfrac38\dfrac{(x-2)^2}{x+1},\, \ln\dfrac e2 +\dfrac12\dfrac{(x-2)^2}{x+2}\right),\quad x\in(2,\infty)\\[4pt] x-1 - \ln x \in\left(-\ln2e +\dfrac18\dfrac{7x^2+20}{x+1},\, -\ln2e +\dfrac{x^2+4}{x+2}\right),\quad x\in(2,\infty)\\ x-3\ln x \in\left(\dfrac{(x-u)(x-7u)}{2u(2x+u)},\dfrac{(x-u)(x+u-6)}{x+u}\right),\quad\text{if}\quad x\in(u,\infty)\\ \end{cases}\tag7$$ where $u$ is given by $(4).$

Assume $X(x)$ and $Y(y)$ monotonic positive functions.

Easily to see that the first condition is satisfied if the function $X(x)$ decreases in the interval $(2,\infty).$

Taking in account $(1),$ the third condition in the form $(7.4)$ means that the production $X(x)Y(y)$ infinitely increases when $x\to\infty.$ Thus, function $Y(y)$ increases in the interval $(1,\infty).$

Taking in account increasing factor $\frac{x-1}x$ in the $\textbf{second condition},$ function $X(x)$ should contain decreasing factor ${\large\frac1x}.$ Taking in account $(7.5),$ it should contain logarithmic factor. It is possible, because the production $x^{-1}\ln^p(x+1)$ decreases in the interval $x\in(1,\infty)$ if $p\in(0,1].$

Looks that the functions $$X(x) = \left(\ln(x+1)\right)^p x^{-q},\quad Y(y)= C\, \left(\ln(x+1)\right)^r(2y)^{-s}\tag8$$ can be solution for certain positive values of $p,q,r,s,$

Also, can be useful function $\dfrac{\ln^p\left(x+q\right)}{x},$ which decreases if $p\in[0,2],\,q\ge \dfrac{223}{168}\, \left(q=\dfrac43\right).$

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