I'll assume you're familiar with the fact that a function is convex if and only if its epigraph is convex.
If the function is positive homogenous, then by just checking definitions, we see that its epigraph is a cone. That is, for all $a > 0$, we have:
$$ \begin{aligned}(\mathbf{x},t) \in \text{epi f} &\Leftrightarrow f(\mathbf{x})\le t \\& \Leftrightarrow af(\mathbf{x})=f(a\mathbf{x}) \le at \\ &\Leftrightarrow (a\mathbf{x},at) \in \text{epi f}\end{aligned}$$
On the other hand suppose the epigraph is a cone. This means for all $a>0$, if $(\mathbf{x},t) \in \text{epi} f$ then $(a\mathbf{x},at) \in \text{epi f}$. Clearly $(\mathbf{x},f(\mathbf{x})) \in \text{epi f}$, so $(a\mathbf{x},af(\mathbf{x})) \in \text{epi f}$, which means that
$$ f(a\mathbf{x}) \le a f(\mathbf{x}).$$
Likewise $(a\mathbf{x},f(a\mathbf{x})) \in \text{epi f}$. So, again, if the epigraph is a cone, $(\mathbf{x},f(a\mathbf{x})/a) \in \text{epi f}$, or:
$$ f(\mathbf{x}) \le f(a\mathbf{x})/a.$$
Combining the two inequalities, we get:
$$ f(\mathbf{x}) \le f(a\mathbf{x})/a \le f(\mathbf{x})$$
Which means $f(a\mathbf{x}) = a f(\mathbf{x})$, ie, the function is positive homogenous.
Since $g$ is non-negative (and $p\geq1$), consider instead the function $h$ defined on $\mathbb{R}$ by
$$\forall x\in\mathbb{R},\ h(x)=\bigl(\max\{0,x\}\bigr)^p$$
With this $h$, we do have:
$$\forall x\in\mathbb{R},\ g(x)^p=(h\circ g)(x),$$
$h$ is convex and we have $\widetilde{h}=h$ is non-decreasing.
Best Answer
If $A$ is a real symmetric matrix, then its maximal eigenvalue is $$\lambda_1(A)=\max_{\|v\|=1} \langle Av, v\rangle \tag1 $$ and its smallest eigenvalues is $$\lambda_n(A)=\min_{\|v\|=1} \langle Av, v\rangle \tag2 $$ For each fixed vector $v$ the function $A\mapsto \langle Av, v\rangle$ is a linear function of $A$. The maximum of any family of linear functions is convex. The minimum of any family of linear functions is concave.