Consider functions $f$ satisfying the following four conditions: (1) $f$ is real-valued and defined for all real numbers. (2) For any two real numbers $x$ and $y$ we have $f(xy)=f(x)f(y)$. (3) For any two real numbers $x$ and $y$ we have $f(x+y) \le 2(f(x)+f(y))$. (4) We have $f(2)=4$. Prove that: a) There is a function $f$ with $f(3)=9$ satisfying the four conditions. b) For any function $f$ satisfying the four conditions, we have $f(3) \le 9$.
Problem
Source: Germany 2022, Problem 6
Tags: function, algebra, algebra proposed, functional equation, Functional inequality, functional inequalities
pi_quadrat_sechstel
25.06.2022 20:09
Tintarn wrote: Consider functions $f$ satisfying the following four conditions: (1) $f$ is real-valued and defined for all real numbers. (2) For any two real numbers $x$ and $y$ we have $f(xy)=f(x)f(y)$. (3) For any two real numbers $x$ and $y$ we have $f(x+y) \le 2(f(x)+f(y))$. (4) We have $f(2)=4$. Prove that: a) There is a function $f$ with $f(3)=9$ satisfying the four conditions. b) For any function $f$ satisfying the four conditions, we have $f(3) \le 9$. a) $f(x)=x^2$ satisfies (1),(2),(3) and (4) since $(x+y)^2\leq 2(x^2+y^2)\Leftrightarrow0\leq(x-y)^2$. We have $f(3)=3^2=9$. b) By (2) we have $f(1)=f(1)^2$ and $f(1)\in\{0,1\}$. By (3) and (4) we have $4=f(2)\leq4f(1)$. Thus $f(1)=1$. Using $f(1)=1,f(2)=2$ and (2) we get $f(2^k)=2^{2k}$. Let $3^n=\sum_{i=0}^{\left\lfloor n\log_23\right\rfloor}a_i2^i$ with $a_i\in\{0,1\}$ and define $M_n$ as the set of indices $i$ with $a_i=1$. Let $M_n=\{b_0,b_1,\ldots,b_m\}$ with $b_0\leq b_1\leq\ldots\leq b_m$. We have \begin{align*} f(3)^n=f\left(3^n\right)=f\left(\sum_{j=0}^m2^{b_j}\right)\leq2f(2^{b_m})+2f\left(\sum_{j=0}^{m-1}2^{b_j}\right)\leq2f(2^{b_m})+4f(2^{b_{m-1}})+4f\left(\sum_{j=0}^{m-2}2^{b_j}\right)\\ \leq\ldots\leq\sum_{j=0}^m2^{m+1-j}f(2^{b_j})=\sum_{j=0}^m2^{m+1-j+2b_j}\leq\sum_{i=0}^{\left\lfloor n\log_23\right\rfloor}2^{\left\lfloor n\log_23\right\rfloor+1+i}\leq\left(\left\lfloor n\log_23\right\rfloor+1\right)2^{2\left\lfloor n\log_23\right\rfloor+1} \end{align*}Thus $f(3)\leq\lim_{n\to\infty}\sqrt[n]{\left(\left\lfloor n\log_23\right\rfloor+1\right)2^{2\left\lfloor n\log_23\right\rfloor+1}}=9$.
laikhanhhoang_3011
14.02.2023 02:56
I think the problem can be solve with only 2 below conditions...
$$\left\{\begin{matrix}f(xy)=f(x)f(y), (i)
& & \\f(x+y) \leq 2\left ( f(x)+f(y) \right ), (ii)
& & \\
\end{matrix}\right.$$Let $x=y$ in $(i),$ we get: $f(x^2)=f(x)^2,$ which means $f$ takes non-negative value at non-negative point.
Also let $x=y=1$ in $(i),$ we get: $f(1)^2=f(1) \rightarrow f(1) \leq 1.$
Define $g(x)=\sqrt{f(x)}, \forall x \geq 0$ and $(c_n):$ $$\left\{\begin{matrix}c_1=\sqrt{2}
& & \\c_{n+1}=\sqrt{\frac{2c_n^2+c_n}{3}}
& & \\
\end{matrix}\right.$$From now, we just consider $x,y \geq 0.$
$\bullet$ Let $x=y$ in $(ii),$ we get: $g(2x) \leq 2g(x).$
Also, $g(1) \leq 1.$
Claim: $g(x+y) \leq c_n.\left ( g(x)+g(y) \right ), \, \forall x,y \geq 0, \forall n \in N^*.$
Proof:
$\,$ $\bullet$ $n=1.$ From $(ii),$ we get: $g(x+y) \leq \sqrt{2\left ( g(x)^2+g(y)^2 \right )} \leq \sqrt{2}\left ( g(x)+g(y) \right ).$
$\,$ $\bullet$ Assume it's true to $n,$ or $g(x+y) \leq c_n.\left ( g(x)+g(y) \right ), \, \forall x,y \geq 0.$
$\,$ Then,
\begin{align*} g\left ( \left ( x+y \right )^2 \right ) &= g(x^2+2xy+y^2) \\
&\leq c_ng(x^2)+c_ng(2xy+y^2) \\
&\leq c_ng(x^2)+c_n\left ( c_ng(2xy)+c_ng(y^2) \right ) \\
& = c_n^2g(2xy)+c_ng(x)^2+c_n^2g(y)^2 \\
&\leq 2c_n^2g(x)g(y)+c_ng(x)^2+c_n^2g(y)^2 , (1).
\end{align*}$\,$ Similarly, we can prove:
\begin{align*} g\left ( \left ( x+y \right )^2 \right ) &\leq 2c_ng(x)g(y)+c_n^2g(x)^2+c_n^2g(y)^2, (2) \\
g\left ( \left ( x+y \right )^2 \right ) &\leq 2c_n^2g(x)g(y)+c_n^2g(x)^2+c_ng(y)^2, (3)
\end{align*}$\,$ Add $(1),$ $(2)$ and $(3)$ up, we get: $3g\left ( \left ( x+y \right )^2 \right ) \leq \left ( 2c_n^2+c_n \right )\left ( g(x)+g(y) \right )^2$
$$\leftrightarrow g(x+y) \leq \sqrt{\frac{2c_n^2+c_n}{3}}.\left ( g(x)+g(y) \right )=c_{n+1}.\left ( g(x)+g(y) \right ),$$$\,$ because $f$ is multiplicative, then also $g.$
q.e.d
Note that, by induction, we can prove $c_n \geq 1, \forall n.$ So $c_{n+1}^2=\frac{2c_n^2+c_n}{3} \leq c_n^2 \rightarrow c_{n+1} \leq c_n.$
Thus, $(c_n)$ is a decreasing sequence, and bounded below by $1.$ So $\textrm{lim}(x_n)=L \geq 1.$ And clearly, $L=1.$
So $$g(x+y) \leq g(x)+g(y), \forall x,y \geq 0$$Note that, $g(1) \leq 1.$ By induction, we can easily prove: $g(n) \leq n, \forall n \in N^* \rightarrow f(n) \leq n^2, \forall n \in N^*.$
In general, $$f(3) \leq 9,$$for any function $f$ satisfying the conditions.
Q.E.D