open new activity based on shared preferances value

HI :) what I'm trying to do is check if user is logged in to the android application and the deleted the application from history and reopend it to make him move to user_menu activity not to the main activity i used shared preferances but it didn't work

in MainActivity i put a function check to check if user is logged in

public class MainActivity extends ActionBarActivity {
    Button login;
    Button reg;
    Button guest;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        check();
    super.onCreate(savedInstanceState);
   }
 public void check()
 {
    SharedPreferences preferences =           PreferenceManager.getDefaultSharedPreferences(this);
    String name = preferences.getString("Name", "");
    if(!name.equalsIgnoreCase(""))
    {
        Intent intent=new Intent(getApplicationContext(),User_Menu.class);
        startActivity(intent); /* Edit the value here*/
    }

and on user_main activity where user go when he log's in i put

public class User_Menu extends ActionBarActivity { AskConnection db;

Button lbsButton;
Button guideButton;
Button duaButton;
Button mediaButton;
Button askButton;
Button azkarButton;
Button logoutButton;
static String nameU=null;
private static final int RESULT_SETTINGS = 1;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_menu);

    //nameU=new String();
/*  setLanguage();
    showUserSettings();*/
    android.app.ActionBar bar = getActionBar();
    bar.setBackgroundDrawable(new ColorDrawable(Color.parseColor("#03A9F4")));

try{

        JSONObject json_data=null;
        for(int i=0;i<LoginConnection.jArray.length();i++){
            json_data = LoginConnection.jArray.getJSONObject(i);
                    nameU=json_data.getString("name");

            //list.add(answer);
        //  print(nameU);

    }

        display();
}


    catch(JSONException e1){
        print("No Data Found");
       } catch (ParseException e1) {
    e1.printStackTrace();
  }