Let $(X,\mathcal B(X),\lambda)$ be some Euclidean space $X\subset \mathbb R^n$ equipped with the Lebesgue measure $\lambda$. Suppose we have a Fredholm integral operator $$T:L^1(X)\to L^1(X):f(\cdot)\mapsto\int_XK(\cdot,y)f(y)\ \mathrm dy,$$ where $K$ is some uniformly bounded kernel, that is not necessarily symmetric. Now suppose that we can find a sequence of kernels $K^m$ converging to $K$ in $L^1(X^2)$ corresponding to Fredholm integral operators $T^m$. It is well-known that if $\rho(T)<1$, where $\rho(T)$ denotes the spectral radius of $T$, that the corresponding Fredholm integral equation of the second kind $$f=g+Tf$$ admits a unique solution for any $g\in L^1$, that the solutions depends continuously on the initial conditions, that the resolvent can be given by a Liouville Neumann series, and furthermore that the resolvent $r$ depends continuously on the initial conditions. In particular, it follows from Corollary 9.3.12 from Gripenberg, Londen, Staffans, Volterra Integral and Functional Equations that for $m$ sufficiently large, $T^m$ admits a resolvent $r^m$, and that $r^m\to r$ in $L^1(X)$.
Can we, somehow, conclude from this that given $\epsilon>0$, we can find some $M$, such that for $m\geq M$, $\rho(T^m)<\rho(T)+\epsilon$? Intuitively, such a results seems to make sense, but I have trouble proving it. It seems that the answer is affirmative when we force the kernel $K$ to be symmetric, see https://arxiv.org/abs/1111.4475. However, in the application I have in mind, I was hoping to do without assumptions like self-adjointness, normality, etc. Also, I was hoping there was maybe a more direct approach.
Any help or reference is much appreciated.