The short answer is no. It's not linear.
Here's the long answer.
There are three kinds of heat transfer: conduction, convection, and radiation.
Conduction means heat transfer due to physical contact. In the case of a building, the conduction is linear-- the amount of energy that leaks into a building is linearly proportional to the difference between the temperature outside and the temperature inside. The constant of proportionality is called the thermal conductivity of the material the heat is passing through. For well-insulated buildings, a great deal of the heat comes in through the windows, as glass is a poor insulator.
Convection means heat transfer due to bulk fluid flow, like wind. Unfortunately, this is hard to model accurately for buildings. It's safe to say that in most places on Earth, for most buildings, convection heats up your building, but not as much as conduction. Still, it screws up your model. It's particularly a problem with cold winds in the winter, where the temperature difference can be higher than in the summer.
For buildings, radiation means the sun. The effects of radiation are roughly proportional to the projected area of your windows over the course of the day. Unfortunately, the contribution to heating from radiation can be greater than that from conduction, especially for a well-insulated house with a lot of windows facing toward the equator. This is where the real nonlinearity comes in-- total irradiance is difficult to measure without considering the geometry of the windows and modeling cloud cover as a function of time.
To add a few numbers-- on a hot night (say, 80 F), you can cool a two-car garage with 1000 W air conditioner (which is actually providing you around 3000 W of cooling, see heat pumps). At noon, the sun averages around 1000 W per square meter, so if your garage has some large windows pointed toward the sun (say, skylights on a roof angled toward the equator), your AC unit can easily be overwhelmed by radiation alone.
The EPSON datasheet has an incorrect formula for calculating the value of the horizontal display width register (HDISP [16h]). The formula should read:
HDISP in number of pixels = ((REG[16h] bits 6-0)) x 8
Instead of:
HDISP in number of pixels = ((REG[16h] bits 6-0)+1) x 8
Best Answer
Mobile phones give quite a good benchmark on display power consumption.
For example a Samsung Galaxy S5 has a 5", 1920x1080 pixels display. And it has a 2800 mAh battery and is able to do a benchmark at 200 nit for around 7.5 hours (based on phonearena).
So the phone needs an average of 375 mA during that time. I'd guess that most of it goes in the display, so there is not that much current to be saved, but they run on a lower voltage, so you can save some power. (If you calculate pixels per milliamp they are insanely more efficient)
It's quite easy to get your hands on those mobile phone displays, but it's a lot harder to get your hands on an efficient driver to use the displays outside of their normal environment (at least that's my experience).
I don't know which interface you want to use, but I think that's part of the problem, going from RPi -> HDMI -> Display is more power hungry than say a display which goes µP -> Display (embedded displayport).
The 3.3 V of the RPi is very weak, you won't be able to power a display directly from it. It's limited to 50 mA if I remember correctly.