Seemingly simple consumer theory problem A function $v: \mathbb{R}^K_+ \xrightarrow{} \mathbb{R}_+$ is is said to be a valuation function if

The value of function $v$ at $x = \textbf{0}$ is $0$: $v(\textbf{0}) = 0$
$v$ is continuous on the domain $\mathbb{R}^K_+$, strictly increasing and strictly concave on $\mathbb{R}^K_{++}$
For any $p \in \mathbb{R}^K_{++}$, the set $A(p) = \{x \in \mathbb{R}^K_+| v(x) \geq p \cdot x\}$ is compact, the set $A^*(p) = A (p)-\{\bf{0}\}$ is nonempty and lies inside $\mathbb{R}^K_{++}$.

Show that a if $v$ is a valuation function, for every $p \in \mathbb{R}^K_{++}$, for every $x \in \mathbb{R}^K_{++}$, there exist $t&gt;0$ small enough that
$$
v(tx) &gt; p\cdot tx
$$
In other word, for every $p&gt;\bf{0}$, if you zoom the set $A(p)$ sufficiently close at the origin, it contains all the points near the origin in first quadrant.
Remark 1: This is the problem I reduce from a microeconomics problem I am working on.  I want to extend the Inada condition to multivariate case in a topological way. My prototype function is $f(x,y)=x^{\alpha}y^{\beta}$ where $\alpha+\beta &lt;1, \alpha&gt;0,\beta&gt;0$.
This family of functions satisfy (1),(2),(3) and the property I want to prove, say (4). I am trying to show either (1),(2),(3) =&gt; (4) or should I include (4) as an additional axiom?
Remark 2: A few examples of $A(p)$

An example of A(p) when p is small



An example of A(p) when p is large


(4) is demonstrated by the geometric fact that the contours approach vertical and horizontal axis near $0$.
It follows indeed from the first three conditions, though I did not find a simple proof. Here is a messy one:
Observe first that, $v(tx)&gt;p\cdot tx$ with $t&gt;0$ is equivalent to $v(tx)/t&gt;px$. By replacing $p$ by some multiple, we can see that the right-side can take any value without changing the left value. The function $t\mapsto v(tx)$ is strictly concave, strictly increasing, and has value $0$ at $0$. The problem is then equivalent to showing that its right derivative is infinite for all $x\gg 0$. That the right derivative exists follows from the concavity of $v$.
As it turns out, it is enough to find a single $x^*\gg0$ such that $t\mapsto v(tx^*)$ has an infinite right derivative at $0$.
To see this, let $x\gg 0$. There is some $\delta&gt;0$ such that $x\gg \delta x^*$. Take some $K&gt;0$. For $t&gt;0$ small enough, we have $v\big(\delta t~x^*\big)/(\delta t)&gt;K/\delta $ or $v(\delta t x^*)/t&gt;K.$
Since $v$ is strictly increasing on the interior,
$v(tx)/t&gt;v(t\delta x^*)/t&gt;K$ for $t&gt;0$ small enough. Therefore, the function $t\mapsto v(tx)$ must have infinite right derivative at $0$.
It remains to find the damn $x^*$. Let $\Delta$ be the $K-1$-dimensional unit simplex. For each $N$, there must exist some $x\in\Delta$ such that the function $t\mapsto v(tx)$ has a right derivative of at least $N$ at $0$. Otherwise, for $p=(N,N,\ldots,N)$ we would have $A(p)-\{0\}=\emptyset.$ Also, $A(p)-\{0\}\subseteq\mathbb{R}^K_{++}$ implies that such an $x$ must be positive in each coordinate. So let $x_n\in\Delta$ be such that $t\mapsto v(tx_n)$ has a right derivative of at least $n$ at $0$. Now let
$$x^*=\sum_{i=1}^\infty\frac{1}{2^i}x_{2^i}.$$
That the relevant series converges follows from any norm being bounded on the compact set $\Delta$ and absolute convergence implying convergence. Clearly, $x^*\gg 0$ since $x_{2^i}\gg 0$ for all $i$. It remains to show that $t\mapsto v(tx^*)$ has an infinite right derivative at $0$. Since
$$t^{-1}~v\bigg(t\sum_{i=1}^\infty\frac{1}{2^i}x_{2^i}\bigg)&gt;t^{-1}~v\bigg(t\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)$$
for all $n$ by $v$ being strictly increasing on the interior, it suffices to show that for each $K&gt;0$, there exists some $n$ such that
$$t^{-1}~v\bigg(t\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)&gt;K$$ for $t&gt;0$ small enough. Let $S_n=\big(\sum_{i=1}^n 1/2^i\big)^{-1}$ and note that
$$t^{-1}v\bigg(tS_n\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)=t^{-1}v\bigg(\sum_{i=1}^n S_n\frac{1}{2^i}t x_{2^i}\bigg)\geq\sum_{i=1}^n S_n\frac{1}{2^i}t^{-1}v(tx_{2^i})$$
by the concavity of $v$. Since $\lim_{t\downarrow 0}t^{-1}v(t x_{2^i})\geq 2^i$, this sum is almost $S_n n$ for $t&gt;0$ small enough. Now $S_n$ converges to $1$, so $S_n n$ is larger than $K$ for $n$ large enough and $t&gt;0$ small enough.