Typescript: generic that extends a type with a generic (higher kinded types)

Unfortunately, typescript does not (yet?) implement higher kinded types.

See https://github.com/Microsoft/TypeScript/issues/1213 for details

This is not exactly what you are looking for but it is as close as I think you can get:

interface Applicative<T> {}

function f<U>(fn: Function, a: U & Applicative<any>): U & Applicative<Function> {

    return null;
}

a will have to be both U (whatever U is) and Applicative<any>. U can not be defined to be a generic type explicitly I am afraid.

Better typing can be achieved by:

function f<U, V>(fn: Function, a: U & Applicative<V>): U & Applicative<Function> { }

I am not entirely sure that the return type in my example is exactly what you want. But you should be able to achieve your required result by adding/changing the required interface on the return type e.g.:

function f<U, V>(fn: Function, a: U & A<V>): U & A<Function>
function f<U, V>(fn: Function, a: U & A<V>): U & A<Function> & B<V>

Or something similar.