Java - Replace interface type for implementation type with bounded type parameters

Question

I'm trying to learn functional programming with Haskell but I'm struggling to understand some concepts. I thought it would be a good idea to translate some of those concepts to Java since it's the language I feel the most comfortable with.

I have the following interface Applicative<A>:

public interface Applicative<A>
{
    public <B> Applicative<B> pure (B value);

    public <B> Applicative<B> apply (Applicative<Function<A, B>> applicative);

    public <B, C> Function<Applicative<B>, Applicative<C>> liftA2 (
        Function<A, Function<B, C>> function);
}

Let's say I have an implementation of the interface ListApplicative<A>. It extends ArrayList<A> and it implements Applicative<A>.

The question is: How do I turn all the Applicative<?> into ListApplicative<?>? Is it even possible? Are there any hacks to do it?

public class ListApplicative<A> extends ArrayList<A> implements Applicative<A>
{
    @Override
    public <B> ListApplicative<B> pure (B value)
    {
        // Implementation
    }

    @Override
    public <B> ListApplicative<B> apply (
        ListApplicative<Function<A, B>> applicative)
    {
        // Implementation
    }

    @Override
    public <B, C> Function<ListApplicative<B>, ListApplicative<C>> liftA2 (
        Function<A, Function<B, C>> function)
    {
        // Implementation
    }
}

I've read a little bit about F-bound types, but I'm not sure what to do. I can't use B or C as parameterized types since their types are unknown until you run the function.

I've looked at other similar questions like this one about functors, this one about generics of generics and this one about interfaces but I didn't find the answer I was looking for.

Thanks a lot!