Let $I$ be a compact interval, say $I:=(0,1)$, and $k\in L^\infty(I\times I)$ a symmetric Markov kernel, i.e. $k(x,y)=k(y,x)$ and $$\int_I k(x,y) d y = 1\qquad\mbox{for almost all } x\in I.$$ Let $K:L^2(I) \to L^2(I)$ be the integral operator $$ K[f](x) := \int_I k(x,y) f(y) d y. $$ Obviously, $K$ is compact and self-adjoint and $1$ is an eigenvalue, the corresponding eigenfunction being a constant. Is there a "general" sufficient condition for this eigenvalue to be simple? In other words, a sufficient condition for the existence of a spectral gap, i.e., that the largest eigenvalue of $K$ on $L^2_0(I)$ is strictly less than one, where $$ L^2_0(I) = \left\{ f\in L^2(I), \int_I f(x) d x = 0 \right\}$$.

I know that by standard theory of Markov chains, the existence of spectral gap is equivalent to geometric ergodicity of $K$. So my question can be reformulated as: Does the nonnegativity, symmetry, stochasticity and boundedness of $k$ already imply geometric ergodicity of $K$, or is this still not enough?

Naively, $K$ has the eigenvalue $1$ on $L^2_0(I)$ if one takes $k(x,y):=\delta(x-y)$, so, obviously, the spectral gap can be as small as one likes (by choosing $k$ a bounded approximation of the Dirac measure), but I still do not see a way to make it equal to zero with a bounded kernel.

Possibly my question is embarassingly trivial for someone familiar with the theory of Markov chains, but I read through some standard references and still haven't found a clear answer to my question: Either "yes", or "no" with a counterexample.