Bifurcation from a characteristic value

Remarks

Every bifurcation point is a characteristic value of $K$: a number $\lambda_0$ with $\varphi = \lambda_0 K \varphi$ for some $\varphi \ne 0$. Indeed, if $\lambda_0$ is not one, then $I - \lambda K$ is invertible with uniformly bounded inverse for $\lambda$ near $\lambda_0$, and $\norm{\varphi} \le C \abs{\lambda} \norm{R(\varphi)}$ forces $\varphi = 0$ for small $\varphi$. Krasnosel'skii's theorem gives a converse: if $R$ is also compact, every characteristic value of $K$ of odd algebraic multiplicity is a bifurcation point.

Example: $X = \R$, $K = 1$ and $R(\varphi) = -\varphi^3$. The characteristic value is $\lambda_0 = 1$, and for $\lambda > 1$ the nonzero solutions are $\varphi = \pm\sqrt{1 - 1/\lambda}$, which tend to $0$ as $\lambda \to 1$; so $1$ is a bifurcation point, a pitchfork bifurcation.

Non-example: $X = \R^2$, $K = I$ and $R(x, y) = (-y^3, x^3)$. The characteristic value $1$ has multiplicity $2$, which is even. The equations are $(1 - \lambda) x + \lambda y^3 = 0$ and $(1 - \lambda) y - \lambda x^3 = 0$; multiplying the first by $y$, the second by $x$ and subtracting gives $\lambda(x^4 + y^4) = 0$. So for $\lambda$ near $1$ the only solution is $x = y = 0$, and $1$ is not a bifurcation point.