Vacuum diode in space charge mode: emission current
Think of the device with zero anode current and both plates at 0V and the the device has been allowed to run a while:

If the inequality holds:

\$ \textrm{If }I_{annode} = 0 \textrm{ then } I_{emmission}+I_{return} =0 \$

this only makes sense if you put a direction on the emission and return currents so one opposes the other. This represents the electrons jumping off the hot cathode, hanging around in between the plates (or bouncing off of one and back to the other), but the net charge on the cathode is going to be zero. 

In circuit you have to have define a current with a direction and a polarity. So in this case I'm going to define all currents as cathode to anode as positive and use the actual electron flow moving in the direction as positive (which is reversed from circuit current nomenclature). But this means that in this operation \$ I_{emission} \$ and \$I_{return} \$ are have opposing signs  on their values

\$I_{emmission} + I_{return} =0 \$ 

Now lets suppose that we increase the voltage on the anode to a positive value:
(Upper diagram). Electrons are heated off the plate (\$ I_{emission} \$) but not all of them come back because some of them hit the anodes metal (its now positive) and flow out of the diode. 

\$I_{emmission}+I_{return} = I_{anode} \$ 

Since there are always electrons bouncing back to the cathode \$ I_{return} \$ is always negative \$ I_{return}&lt;0\$ (in the way I've defined things). With the author including the inequality \$ I_{return}&lt;0\$, it means that the return current is defined as going from cathode to anode and is always negative.  

The important takeaway is that with a negative return current you can't have more anode current than emission current.
\$I_{emmission}&gt; I_{anode} \$ 
 And that means you'll have to keep the cathode sufficiently hot.  



Diagrams from and further reading here