Kernel feature and derivative of kernel feature linearly independent?

Suppose we have a strictly positive definite symmetric kernel $k$ on an open set $\Omega\subset\mathbb R$. By "strictly" I mean that all kernel matrices $(k(x_i,x_j))_{i,j}$ with distinct $x_i$ are positive definite (not only semi-definite). Let us further assume that $k$ is $C^1$.

Conjecture: For $x_1,x_2\in\Omega$, the two functions $k(\cdot,x_1)$ and $\partial_1k(\cdot,x_2)$ are linearly independent.

Here, $\partial_1$ denotes the derivative w.r.t. the first component.

In fact, I can prove this in a special case: assume that $\Omega=\mathbb R$ and $k(x,y) = \phi(x-y)$ with a positive definite function $\phi$. In this case, due to Bochner's theorem, we have $\phi(x) = \int e^{ixz}\,d\mu(z)$ with a finite measure $\mu$ on $\mathbb R$. Assume furthermore that $\mu$ is absolutely continuous, i.e., $d\mu(z) = \rho(z)\,dz$ with $\rho\in L^1(\mathbb R)$. The symmetry and strict definiteness of $k$ then imply that $\rho$ is even and has full support. Now, $ak(\cdot,x_1) + b\partial_1k(\cdot,x_2) = 0$ means$$a\int e^{i(x-x_1)z}\rho(z)\,dz - ib\int ze^{i(x-x_2)z}\rho(z)\,dz = 0\qquad\forall x\in\mathbb R.$$Hence, with $f(z) = (ae^{-ix_1z}-ibze^{-ix_2z})\rho(z)$ this gives $\mathcal F^{-1}f = 0$, where $\mathcal F$ is the Fourier transform. Thus, $f(z)=0$ for all $z\in\mathbb R$. As $\rho$ has full support, we conclude that $ae^{-ix_1z}-ibze^{-ix_2z}=0$ for all $z\in\mathbb R$. Setting $z=0$ yields $a=0$, and $b=0$ follows from setting $z=1$.

EDIT. What I find interesting is the fact that the above reasoning does not work for $\Omega\neq\mathbb R$. Because in this case, we only have $\mathcal F^{-1}f(x)=0$ for all $x\in\Omega$, so we cannot conclude that $f=0$. However, the claim is still true if $k\in C^2$. To see this, consider the RKHS norm:\begin{align}0&= \|ak(\cdot,x_1) + b\partial_1k(\cdot,x_2)\|^2 = a^2k(x_1,x_1) + 2ab\partial_1k(x_1,x_2) + b^2\partial_1\partial_2k(x_2,x_2)\\&= \int (a^2 + 2abize^{i(x_1-x_2)z} + b^2z^2)\rho(z)\,dz\\&= \int|a-ibze^{-i(x_1-x_2)z}|^2\rho(z)\,dz,\end{align}from which it follows that $a-ibze^{-i(x_1-x_2)z} = 0$ for all $z\in\mathbb R$, leading to $a=b=0$ as above.