If $g(x+y)\le2\cdot\max\{g(x),g(y)\}$ and $g(xy)=g(x)g(y)$, show that $g(x+y)\le g(x)+g(y)$. 
Let $g$ be a multiplicative function from $\mathbb R^+$ to $\mathbb R^+$. If $g(x+y)\le2\cdot\max\{g(x),g(y)\}$ for all positive reals $x,y$, prove that $g$ is subadditive.

For the function $g$, being multiplicative means $g(xy)=g(x)g(y)$ for all $x,y\in\mathbb R^+$, and being subadditive means $g(x+y)\le g(x)+g(y)$ for all $x,y\in\mathbb R^+$.
Edit: My understanding:

*

*$g$ is multiplicative so it's a "mixture" of power functions, i.e., given any positive real $u\ne1$, there is a corresponding dense subset $S_t=\{u^r\mid r\in\mathbb Q\}$ of $\mathbb R^+$ on which $g(x)=x^t$ where $t$ is fixed such that $g(u)=u^t$.

*If the exponents of these power functions are all less than or equal to $1$ (i.e., if $g(x)\ge x$ for all $0<x\le1$ and $g(x)\le x$ for all $x>1$), then $g$ is obviously subadditive.

*If there exists a power function with exponent greater than $1$, then I can show there exist power functions with exponents arbitrarily small and arbitrarily large. Clearly the existence of a non-subadditive pair implies the existence of this power function, but I'm not sure the existence of this power function on its own is enough to reach a contradiction.

*I feel like a proof by contradiction is the way to go, but I haven't been able to construct $x',y'$ given $g(x+y)>g(x)+g(y)$ such that $x'+y'=x+y$ and $g(x+y)>2\cdot\max\{g(x'),g(y')\}$.

Progress update:

*

*Suppose $g(a)+g(1-a)<1$ (we lose no generality assuming $a+b=1$). Then I can show $g\left(a^2\right)+g\left(1-a^2\right)\le\bigl(f(a)+f(1-a)\bigr)^2<1$ or $g\left(\frac{2a}{1+a}\right)+g\left(1-\frac{2a}{1+a}\right)<1$. In the first case, clearly we have constructed a sum that is smaller than the original, but in the second case I'm not sure. If somehow we can repeat this process until we obtain a sum smaller than $\frac12$, we get the desired contradiction.

 A: First, consider the following lemma.

Assume that $ g : \mathbb R ^ + \to \mathbb R ^ + $ is a function such that for some constant $ c \in \mathbb R ^ + $ we have
$$ g ( x + y ) \le c \bigl ( g ( x ) + g ( y ) \bigr ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R ^ + $. Then
$$ g ( x + y + z ) \le \frac { 2 c ^ 2 + c } 3 \bigl ( g ( x ) + g ( y ) + g ( z ) \bigr ) \tag 1 \label 1 $$
for all $ x , y , z \in \mathbb R ^ + $.

To prove the lemma, use \eqref{0} twice to get
$$ g ( x + y + z ) \le c g ( x ) + c ^ 2 g ( y ) + c ^ 2 g ( z ) $$
for all $ x , y , z \in \mathbb R ^ + $. Then, apply cyclic permutations to $ ( x , y , z ) $ in the above inequality and conclude
$$ g ( y + z + x ) \le c g ( y ) + c ^ 2 g ( z ) + c ^ 2 g ( x ) $$
and
$$ g ( z + x + y ) \le c g ( z ) + c ^ 2 g ( x ) + c ^ 2 g ( y ) \text . $$
Finally, add up the three inequalities to get \eqref{1}.

Back to the original problem, consider $ g : \mathbb R ^ + \to \mathbb R ^ + $ satisfying
$$ g ( x y ) = g ( x ) g ( y ) \tag 2 \label 2 $$
and
$$ g ( x + y ) \le 2 \max \{ g ( x ) , g ( y ) \} \tag 3 \label 3 $$
for all $ x , y \in \mathbb R ^ + $. Putting $ x = y = 1 $ in \eqref{2} and noting that $ f ( 1 ) > 0 $, we get $ g ( 1 ) = 1 $. Then, setting $ x = y = 1 $ in \eqref{3} we have $ f ( 2 ) \le 2 $.
Define the sequence $ ( c _ n ) _ { n = 0 } ^ \infty $ of positive real numbers with $ c _ 0 = 2 $ and $ c _ { n + 1 } = \sqrt { \frac { 2 c _ n ^ 2 + c _ n } 3 } $ for all nonegative integers $ n $. It's straightforward to inductively verify $ c _ n > 1 $ and $ c _ { n + 1 } < c _ n $ for all nonegative integers $ n $. Then, note that $ t \mapsto \frac { \frac 2 3 t + 1 } { \sqrt { \frac { 2 t ^ 2 + t } 3 } + 1 } $ is strictly decreasing on the interval $ [ 1 , 2 ] $, and thus for any nonnegaive integer $ n $,
\begin{align*}
c _ { n + 1 } - 1
& = \sqrt { \frac { 2 c _ n ^ 2 + c _ n } 3 } - 1 \\
& = \frac { \frac { 2 c _ n ^ 2 + c _ n } 3 - 1 } { \sqrt { \frac { 2 c _ n ^ 2 + c _ n } 3 } + 1 } \\
& = \frac { \frac 2 3 c _ n + 1 } { \sqrt { \frac { 2 c _ n ^ 2 + c _ n } 3 } + 1 } ( c _ n - 1 ) \\
& < \frac { \frac 2 3 \times 1 + 1 } { \sqrt { \frac { 2 \times 1 ^ 2 + 1 } 3 } + 1 } ( c _ n - 1 ) \\
& = \frac 5 6 ( c _ n - 1 ) \text ,
\end{align*}
which implies $ c _ n - 1 \le \left ( \frac 5 6 \right ) ^ n $, and consequently
$$ \lim _ { n \to \infty } c _ n = \inf _ { n = 0 } ^ \infty c _ n = 1 \text . $$
Now, letting $ y = x $ in \eqref{2} we get
$$ g \left ( x ^ 2 \right ) = g ( x ) ^ 2 \tag 4 \label 4 $$
for all $ x \in \mathbb R $. Note that \eqref{3} implies \eqref{0} with $ c = c _ 0 $. Now, assume that for some nonnegative integer $ n $ we have \eqref{0} with $ c = c _ n $ and all $ x , y \in \mathbb R ^ + $, and use \eqref{2}, \eqref{4} and the above lemma to observe that
\begin{align*}
g ( x + y ) ^ 2
& = g \left ( ( x + y ) ^ 2 \right ) \\
& = g \left ( x ^ 2 + 2 x y + y ^ 2 \right ) \\
& \le c _ { n + 1 } ^ 2 \Bigl ( g \left ( x ^ 2 \right ) + g \left ( 2 x y \right ) + g \left ( y ^ 2 \right ) \Bigr ) \\
& = c _ { n + 1 } ^ 2 \left ( g ( x ) ^ 2 + g ( 2 ) g ( x ) g ( y ) + g ( y ) ^ 2 \right ) \\
& \le c _ { n + 1 } ^ 2 \left ( g ( x ) ^ 2 + 2 g ( x ) g ( y ) + g ( y ) ^ 2 \right ) \\
& = c _ { n + 1 } ^ 2 \bigl ( g ( x ) + g ( y ) \bigr ) ^ 2 \text ,
\end{align*}
which implies \eqref{0} with $ c = c _ { n + 1 } $. Therefore, we inductively get for all nonnegative integers $ n $ \eqref{0} with $ c = c _ n $, and consequently with $ c = \inf _ { n = 1 } ^ \infty c _ n = 1 $, which completes the proof.
A: First consider the following lemmas.

Assume that $ h : \mathbb Z ^ + \to \mathbb R ^ + $ is a function satisfying $ h ( 1 ) = 1 $ and
$$ h ( j + k ) \le 2 \max \{ h ( j ) , h ( k ) \} \tag 0 \label {eqn0} $$
for all $ j , k \in \mathbb Z ^ + $. Then $ h ( k ) < 2 k $ for all $ k \in \mathbb Z ^ + $.

To see why, first let $ j = k $ in \eqref{eqn0} to get $ h ( 2 k ) \le 2 h ( k ) $ for all $ k \in \mathbb Z ^ + $. Use this together with $ h ( 1 ) = 1 $ to inductively get $ h ( 2 ^ l ) \le 2 ^ l $ for all nonnegative integers $ l $. Now use strong induction on $ k $ to prove $ h ( k ) < 2 k $: in case $ k = 2 ^ l $ for some nonnegative integer $ l $, we already have $ h ( k ) \le k < 2 k $. In case $ k $ is not of that form, there are $ l , r \in \mathbb Z ^ + $ such that $ k = 2 ^ l + r $ and $ r < 2 ^ l $. Note that we have $ 2 ^ l < k $ hence $ 2 ^ { l + 1 } < 2 k $, and also $ r < \frac k 2 $ hence $ 4 r < 2 k $. Therefore, using \eqref{eqn0} and the induction hypothesis $ h ( r ) < 2 r $, we have
$$ h ( k ) = h \left ( 2 ^ l + r \right ) \le 2 \max \left \{ h \left ( 2 ^ l \right ) , h ( r ) \right \} \le 2 \max \left \{ 2 ^ l , 2 r \right \} = \max \left \{ 2 ^ { l + 1 } , 4 r \right \} < 2 k \text . $$

Assume that $ g : \mathbb R ^ + \to \mathbb R ^ + $ is a function such that for some constant $ c \in [ 1 , + \infty ) $, we have
$$ g ( x + y ) \le c \bigl ( g ( x ) + g ( y ) \bigr ) \tag 1 \label {eqn1} $$
for all $ x , y \in \mathbb R ^ + $. Then for all nonegative integers $ n $ and all $ x _ 0 , \dots , x _ n \in \mathbb R ^ + $, we have
$$ g \left ( \sum _ { i = 0 } ^ n x _ i \right ) \le c ^ { \lceil \log _ 2 ( n + 1 ) \rceil } \sum _ { i = 0 } ^ n g ( x _ i ) \text . \tag 2 \label {eqn2} $$

Use strong induction on $ n $ for a proof. For $ n = 0 $, it is trivial. For $ n > 0 $, use \eqref{eqn1} and the induction hypothesis for $ \left \lfloor \frac n 2 \right \rfloor $ and $ \left \lceil \frac n 2 \right \rceil - 1 $ to see that
\begin{align*}
g \left ( \sum _ { i = 0 } ^ n x _ i \right )
& = g \left ( \sum _ { i = 0 } ^ { \left \lfloor \frac n 2 \right \rfloor } x _ i + \sum _ { i = \left \lfloor \frac n 2 \right \rfloor + 1 } ^ n x _ i \right ) \\
& \le c g \left ( \sum _ { i = 0 } ^ { \left \lfloor \frac n 2 \right \rfloor } x _ i \right ) + c g \left ( \sum _ { i = \left \lfloor \frac n 2 \right \rfloor + 1 } ^ n x _ i \right ) \\
& \le c ^ { \left \lceil \log _ 2 \left ( \left \lfloor \frac n 2 \right \rfloor + 1 \right ) \right \rceil + 1 } \sum _ { i = 0 } ^ { \left \lfloor \frac n 2 \right \rfloor } g ( x _ i ) + c ^ { \left \lceil \log _ 2 \left \lceil \frac n 2 \right \rceil \right \rceil + 1 } \sum _ { i = \left \lfloor \frac n 2 \right \rfloor + 1 } ^ n g ( x _ i ) \\
& \le c ^ { \left \lceil \log _ 2 \left ( \left \lfloor \frac n 2 \right \rfloor + 1 \right ) \right \rceil + 1 } \sum _ { i = 0 } ^ { \left \lfloor \frac n 2 \right \rfloor } g ( x _ i ) + c ^ { \left \lceil \log _ 2 \left ( \left \lfloor \frac n 2 \right \rfloor + 1 \right ) \right \rceil + 1 } \sum _ { i = \left \lfloor \frac n 2 \right \rfloor + 1 } ^ n g ( x _ i ) \\
& = c ^ { \left \lceil \log _ 2 \left ( \left \lfloor \frac n 2 \right \rfloor + 1 \right ) \right \rceil + 1 } \sum _ { i = 0 } ^ n g ( x _ i ) \\
& = c ^ { \left \lceil \log _ 2 ( n + 1 ) \right \rceil } \sum _ { i = 0 } ^ n g ( x _ i ) \text .
\end{align*}

Back to the original problem, consider $ g : \mathbb R ^ + \to \mathbb R ^ + $ satisfying
$$ g ( x y ) = g ( x ) g ( y ) \tag 3 \label {eqn3} $$
and
$$ g ( x + y ) \le 2 \max \{ g ( x ) , g ( y ) \} \tag 4 \label {eqn4} $$
for all $ x , y \in \mathbb R ^ + $. Note that putting $ x = y = 1 $ in \eqref{eqn3} we get $ g ( 1 ) = 1 $, and by \eqref{eqn3} and mathematical induction we have
$$ g ( x ^ n ) = g ( x ) ^ n $$
for all $ x \in \mathbb R ^ + $ and all nonnegative integers $ n $. As \eqref{eqn4} implies \eqref{eqn1} with $ c = 2 $, by the second lemma, we have \eqref{eqn2} with $ c = 2 $ for all nonnegative integers $ n $ and all $ x _ 0 , \dots , x _ n \in \mathbb R ^ + $. Letting $ h $ be the restriction of $ g $ to $ \mathbb Z ^ + $, \eqref{eqn0} is satisfied, and therefore by the first lemma, $ g ( k ) < 2 k $ for all $ k \in \mathbb Z ^ + $. In particular, we have this for $ k = { n \choose i } $ for all nonnegative integers $ n $ and $ i $ with $ i \le n $. Combining all these, consider any $ x , y \in \mathbb R ^ + $ and observe that for any nonnegative integer $ n $,
\begin{align*}
g ( x + y ) ^ n
& = g \bigl ( ( x + y ) ^ n \bigr ) \\
& = g \left ( \sum _ { i = 0 } ^ n { n \choose i } x ^ i y ^ { n - i } \right ) \\
& \le 2 ^ { \left \lceil \log _ 2 ( n + 1 ) \right \rceil } \sum _ { i = 0 } ^ n g \Biggl ( { n \choose i } x ^ i y ^ { n - i } \Biggr ) \\
& = 2 ^ { \left \lceil \log _ 2 ( n + 1 ) \right \rceil } \sum _ { i = 0 } ^ n g \Biggl ( { n \choose i } \Biggr ) g \left ( x ^ i \right ) g \left ( y ^ { n - i } \right ) \\
& < 2 ^ { \left \lceil \log _ 2 ( n + 1 ) \right \rceil } \sum _ { i = 0 } ^ n 2 { n \choose i } g ( x ) ^ i g ( y ) ^ { n - i } \\
& = 2 ^ { \left \lceil \log _ 2 ( n + 1 ) \right \rceil + 1 } \bigl ( g ( x ) + g ( y ) \bigr ) ^ n \text .
\end{align*}
This means that we must have
$$ g ( x + y ) \le \sqrt [ n ] { 2 ^ { \left \lceil \log _ 2 ( n + 1 ) \right \rceil + 1 } } \bigl ( g ( x ) + g ( y ) \bigr ) $$
for all $ n \in \mathbb Z ^ + $, which implies $ g ( x + y ) \le g ( x ) + g ( y ) $, since $ \inf _ { n \in \mathbb Z ^ + } \sqrt [ n ] { 2 ^ { \left \lceil \log _ 2 ( n + 1 ) \right \rceil + 1 } } = 1 $.
