A diode optocoupler can be used as a negative current source. However, you will not get more than -0.6 V. The transfer coefficient of such an optocoupler may be 1%. By setting a current of 5 mA through the input diode, you will get a current of approximately 50 µA. This is more than enough for a high impedance load. -50µA*10kOhm=-0.5V.

 

I finally have come to the conclusion that the only way to keep linearity is to create a split supply. I have ordered a center tapped transformer to do this.

You can use a switching regulator to make a negative supply, or you can use half wave rectifiers on a transformer with no centertap.

Here's an example.