Typescript Generics and creating new instances

Your clue is in the definition of deserializeObject. You can't use a type in an expression. Types are erased after compilation T will not exist at runtime. That is why deserializeObject has a second parameter that takes in the class as a constructor signature (the classReference parameter).

You can do the same, and take the class as an extra parameter:

export class JsonSerializable {
    toJson(): any {
        const jsonConvert = new JsonConvert();
        jsonConvert.valueCheckingMode = ValueCheckingMode.ALLOW_NULL;
        return jsonConvert.serialize(this);
    }

    static fromJson<T extends JsonSerializable>(data: any, classReference: {
        new (): T;
    }): T {
        const jsonConvert = new JsonConvert();
        jsonConvert.valueCheckingMode = ValueCheckingMode.ALLOW_NULL;

        return jsonConvert.deserializeObject(data, classReference);
    }
}

class MyClass extends JsonSerializable {
    public f = ""
}

let a = JsonSerializable.fromJson({}, MyClass); // a is MyClass

Or if you don't mind invoking fromJson on the class that you want deserialize, you can use this (with an appropriate type annotation on for the this parameter):

export class JsonSerializable {
    toJson(): any {
        const jsonConvert = new JsonConvert();
        jsonConvert.valueCheckingMode = ValueCheckingMode.ALLOW_NULL;
        return jsonConvert.serialize(this);
    }

    static fromJson<T extends JsonSerializable>(this: new () => T, data: any): T {
        const jsonConvert = new JsonConvert();
        jsonConvert.valueCheckingMode = ValueCheckingMode.ALLOW_NULL;

        return jsonConvert.deserializeObject(data, this);
    }
}

class MyClass extends JsonSerializable {
    public f = ""
}

let a = MyClass.fromJson({}); // a is MyClass