TITLE: Find an optimizer for $g(x,y)$ if it exists
QUESTION [0 upvotes]: Consider $f:\mathbb{R}^{2}_{0} \to \mathbb{R}_{0}$ such that $f(x,y)$ is a continuous function and satisfies the following properties:

$f(x,y) = f(y,x)$
$f(tx,ty) = tf(x,y) \ \forall \ t > 0 $
$f(1,1) = 1$

Can we show that if $g(x,y) := 3f(x,y) - 2(x+y)$, then $\underset{x,y}{\text{argmax}}[g(x,y)] = (0,0)$ assuming a maximizer exists?
I can only show that $g(x^{*},y^{*}) =0$ since otherwise $g(2x^{*},2y^{*}) > g(x,y)$ yields a contradiction if $g(x^{*}, y^{*}) \neq 0$. I can neither think of a counter-example nor a proof to complete the solution.
Note that $f$ is not necessarily (partially) differentiable. $\mathbb{R}_0$ is the set of non-negative reals.

REPLY [0 votes]: By conditions 2 and 3, $f(t,t)=t$ for all $t>0$. By continuity, $f(0,0)=\lim_{t\to 0^+} f(t,t)=\lim_{t\to 0^+}t=0$.  From this it follows that $g(0,0)=0$.
As you said, if $g$ has a maximizer, then the maximum value must be $0$. We also know that $g(0,0)=0$. So if we're interested in showing that $(0,0)$ is a maximizer, we're done. If we want to show that $(0,0)$ is the unique maximizer, then I think the result is not true. That is, we can have other maximizers.
Let $p=1/\log_2(3/2)\approx 1.709511$. For non-negative real $x,y$, define $$f(x,y)=\Bigl(\frac{|x|^p+|y|^p}{2}\Bigr)^{1/p}.$$  This function is continuous and satisfies conditions $1$-$3$.
Note that $$g(0,1)=3/2^{1/p}-2=3/2^{\log_2(3/2)}-2=3/(3/2)-2=0.$$ By symmetry and positive homogeneity, $g(t,0)=g(0,t)=0$ for all $t\geqslant 0$.
We will show that $g(x,y)<0$ for all $(x,y)$ with $x,y>0$. This will show that $g$ has a maximizer (and in fact, the set of all $(x,y)\in \mathbb{R}_0^2$ with $\min \{x,y\}=0$ is the exact set of maximizers). Fix $x,y>0$. Define $t=x+y$ and $\theta=\frac{x}{x+y}$. So $$\theta(t,0)+(1-\theta)(0,t)=\frac{x}{x+y}(x+y,0) + \frac{y}{x+y}(0,x+y)=(x,y).$$ That is, $(x,y)$ is a convex combination of the points $(t,0)$ and $(0,t)$ with convex coefficients of $\theta$ and $1-\theta$. Note that $0<\theta, 1-\theta<1$.
Next we note that $$f(\theta (t,0)+(1-\theta)(0,t))<\theta f(t,0)+(1-\theta)f(0,t).$$ This follows from strict convexity of the $L_p$ norm for $1<p<\infty$ (see this answer). Therefore $$g(x,y)=3f(\theta (t,0)+(1-\theta)(0,t)) - 2(x+y) < 3\theta f(t,0)+3(1-\theta)f(0,t) - 2\theta t-2(1-\theta)t = \theta g(t,0)+(1-\theta)g(0,t)=0.$$
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