C# – Unity [dependency] injection and Inheritance

My question is as follows:
I have a base controller (ASP.Net MVC controller) called ApplicationController, and I want all my controller to inherit from it. this base controller has a ILogger property, marked with a [Dependency] attribute. (yes, I know I should use constructor injection, I'm just curious about this attribute).

I created the container, registered types, changed the default factory, everything is fine. the problem is that when I try to use my Logger property in the derived controller, it's not resolved.

what am I doing wrong? why doesn't the container resolves the base class dependencies when creating the derived controller?

code samples:

ApplicationController:

public class ApplicationController : Controller
{
    [Dependency]
    protected ILogger _logger { get; set; }

}

derived controller:

public class HomeController : ApplicationController
{
    public HomeController()
    {

    }
    public ActionResult Index()
    {
        _logger.Log("Home controller constructor started.");
        ViewData["Message"] = "Welcome to ASP.NET MVC!";

        return View();
    }

    public ActionResult About()
    {
        return View();
    }
}

Unity controller factory:

public class UnityControllerFactory : DefaultControllerFactory
{
    private readonly IUnityContainer _container;
    public UnityControllerFactory(IUnityContainer container)
    {
        _container = container;
    }

    protected override IController GetControllerInstance(Type controllerType)
    {
        return _container.Resolve(controllerType) as IController;
    }
}

Global.asax.cs sample:

protected void Application_Start()
    {
        _container = new UnityContainer();
        _container.RegisterType<ILogger, Logger.Logger>();
        UnityControllerFactory factory = new UnityControllerFactory(_container);
        ControllerBuilder.Current.SetControllerFactory(factory);

        RegisterRoutes(RouteTable.Routes);
    }

I'm quite new to Unity, so maybe I did something wrong.

thanks,
Ami.