5

Let $a, b \in \mathbb{R}, 0 \lt a \lt b$ and $f:\mathbb{R} \rightarrow \mathbb{R}$ such that: $$f(x^2 +ay) \ge f(x^2 +by), \forall x,y \in \mathbb{R} \tag1$$

Prove $f$ is constant on $(0, \infty)$

I don't know how to start, any idea is appreciated.

Playing a little bit with (1) I can get:

For $y \lt 0, x= \sqrt {-y}$ from (1) $f(0) \ge f((b-a)y), \forall y \lt 0$ or, similar, $f(0) \ge f((a-b)y), \forall y \gt 0$ also $f(0) \ge f(y) \forall y \lt 0$

but it doesn't seem to be helpful.

4 Answers4

2

Let $0<z_1 < z_2$ such that $az_2\leq bz_1$. Then for: \begin{align} \begin{cases} x^2&=\left(z_2-\frac{b}{a}z_1\right)/\left(1-\frac{b}{a}\right)\\ y&=\left( z_1 - x^2\right)/a \end{cases} \end{align} We have: $f(z_1) \geq f(z_2)$. For \begin{align} \begin{cases} x^2 &= \left(z_2-\frac{a}{b}z_1\right)/\left(1-\frac{a}{b}\right)\\ y &= \left( z_1 -x^2\right)/b \end{cases} \end{align} We have $f(z_2)\geq f(z_1)$. Hence $f(z_1)=f(z_2)$. Since $z_1 $ and $z_2$ were arbitrary we get $f(z)=C$ for all $z>0$.

Edit. Take $z_1 =1$ then we know that $$\{C\}=f\left(\bigg[1, \frac ba\bigg]\right)=f\left(\bigg[\frac ba, \frac {b^2}{a^2}\bigg]\right)=....$$ and this gets all numbers in $[1,\infty)$ since $(b/a)^n\to \infty$ as $n\to \infty$.

Now take $z_2=1$ then we know that: \begin{align} \{C\}= f\left(\bigg[\frac ab,1\bigg]\right)=f\left(\bigg[\frac {a^2}{b^2},\frac ab\bigg]\right)=.... \end{align} And this gets all numbers in $(0,1]$. This proves the result since $(a/b)^n\to 0$ as $n\to \infty$.

I'll leave the small details to you.

Shashi
  • 8,738
1

Suppose we have that $f(x_1)>f(x_2)$ for some $x_1<x_2$, then if we set $$x=\sqrt{\frac{bx_2-ax_1}{b-a}}\qquad\text{and}\qquad y=\frac{x_2-x_1}{a-b}$$ we reach a contradiction. Thus we have that $f(x_1)\le f(x_2)$ for all $x_1\le x_2$. It then follows that $f(x^2+by)\ge f(x^2+ay)$ when $x,y\in\mathbb{R^{+}}$ and hence $$f(x^2+ay)=f(x^2+by)\qquad x,y\in\mathbb{R}^{+}$$ At which point the result is fairly easy to derive.

Will Fisher
  • 5,112
1

If $z, w > 0$ are such that $z/w \in (\frac{a}{b}, \frac{b}{a})$, then

$$ \begin{cases} x^2 = \frac{bz-aw}{b-a} \\ y = \frac{w-z}{b-a} \end{cases} \quad\Rightarrow\quad f(z) = f(x^2+ay) \geq f(x^2+by) = f(w) $$

and switching the role of $z$ and $w$ gives the reverse inequality $f(w) \geq f(z)$. So the inequalities are saturated and we get $f(z) = f(w)$. Now writing $\alpha = b/a$, what we have shown is rephrased as:

$$ \forall s, t \in \mathbb{R} \ : \ |s - t| < 1 \quad \Rightarrow \quad f(\alpha^s) = f(\alpha^t). $$

From here it is easy to deduce that $f(\alpha^s) = f(\alpha^t)$ for all $s, t \in \mathbb{R}$, from which the claim follows.

Sangchul Lee
  • 167,468
  • Maybe $x^2 = \frac{bz-aw}{b-a}$? –  Dec 16 '17 at 18:14
  • @EugenCovaci, Ah yes, you are right. Fixed! – Sangchul Lee Dec 16 '17 at 18:15
  • Yes, but you need $bz-aw \ge 0$, you cannot switch just like that. Anyway, I think your idea can work somehow. –  Dec 16 '17 at 18:17
  • @EugenCovaci, That is why I am assuming $z/w \in (a/b, b/a)$. We have both $bz-aw > 0$ and $bw-az>0$. – Sangchul Lee Dec 16 '17 at 18:18
  • This case you cannot get $f$ constant for all (0, \infty), I think. I'm worried about the fact that $z, w$ are not arbitrary. –  Dec 16 '17 at 18:20
  • @EugenCovaci, That is why we need an extra step as in my answer. My assumption $z/w \in (a/b, b/a)$ tells that, if we let $\alpha = b/a$ and $(z, w) = (\alpha^s, \alpha^t)$, then whenever $s, t$ are real and $|s - t| < 1$ we have $f(\alpha^s) = f(\alpha^t)$. For arbitrary $s, t \in \mathbb{R}$, you can find a natural number $n$ so that $|s-t|/n < 1$. Then letting $s_k = s + (k/n)(t-s)$ shows that $f(\alpha^{s_k}) = f(\alpha^{s_{k+1}})$ for all $k = 0, \cdots, n-1$, from which we obtain $f(\alpha^s) = f(\alpha^{s_0}) = \cdots = f(\alpha^{s_n}) = f(\alpha^t)$. – Sangchul Lee Dec 16 '17 at 18:23
1

Making the change of coordinates $$\begin{pmatrix} 1 & a \\ 1 & b\end{pmatrix}\begin{pmatrix}x^2 \\ y\end{pmatrix} = \begin{pmatrix}X \\ Y\end{pmatrix}$$ we find that the given condition can be rephrased as $$ f(X)≥f(Y), \qquad \text{where }X,Y\in\mathbb R, \ bX -aY ≥ 0 $$ where the second condition can be written $ X≥\frac{aY}{b}$. So actually we have $f(X)≥f(Y)$ if either $X≥Y$ or if $Y≥X≥\frac{aY}{b}$. Hence, we conclude that for any fixed $Y$, we have $$ f(X) ≥ f(Y) \text{ for all } X\in[\tfrac{aY}{b} ,\infty) $$ and for any fixed $X$, $$f(X) ≥ f(Y) \text{ for all } Y\in(-\infty ,\tfrac{bX}{a}] $$ Relabelling $X\leftrightarrow Y$ in this second inequality gives us $$f(X) = f(Y) \text{ for all } X \in [\tfrac{aY}{b}, \tfrac{bY}{a}]$$ By choosing the two-sided sequence of $Y$-points, $n\in\mathbb Z$, $$Y_n=\frac{a^n}{b^n}\to\begin{cases}\infty, & \text{if }\ n\to -\infty \\ 0, &\text{if }\ n\to \infty\end{cases}$$ we conclude that $f$ is constant on $(0,\infty)$.

Calvin Khor
  • 34,903