Is single tree shadow locally affecting air temperature? In other words, is the air temperature under a single tree different from a couple meters away, in a hot day and under the sun?
There are several effects potentially interacting:


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*Tree transpiration cools the air around the tree and the surface under the tree in the shadow doesn't heat the air above it.

*Air temperature stays uniform under and around the tree because of constant air movement.

 A: If you measure with sufficient accuracy, you will see a difference. Of course, any object you use to measure the air temperature might be exposed to sunlight and therefore experience significant difference in "apparent" temperature; but if we assume that you have a well shielded device that admits air but is insensitive to the effects of radiated heat (from the sun above and the soil below), there are a couple of effects in play.
First - the ground in the shade will be cooler than the ground in the sun. This will set up a convection current - the air in the shade close to the ground will cool down, and become more dense (A). This denser air will then push sideways, displacing the hot air close to the ground on the sunny side (B). That air, in turn, will rise (C) - and it will be pulled (D) into the shaded region since air was leaving at ground level, and had to be replaced with air above it (E). This is called a convection cell - schematically it looks like this:

This tends to change the answer depending on exactly where you are measuring the temperature, since there is a permanent "mixing" of the air going on, which is driven by temperature differences. Definitely, close to the ground this would show a lower temperature.
The second effect is evaporation. When moisture evaporates, it extracts heat from the liquid (heat of evaporation) and therefore the liquid itself cools down a little bit. At the very top of the tree, the leaves will be very hot because of the sunlight, and that complicates things; but lower down, the leaves will be in the shade. A leaf in the shade that is evaporating water will cool down a little, but the water vapor will displace the air and make it less dense. So it's a fair question to ask - would that air rise (since it's less dense) or fall (since it's colder). Let's do that calculation.
Assume I evaporate $m$ grams of water, and this affects the temperature / density of a volume $V$ (initial mass $M$) of air. The heat of evaporation is approximately 2300 kJ/kg, and the heat capacity of air is approximately 1 kJ/kg/K. For this example, the air temperature will change by $2300\frac{m}{M}$. Now if $m$ water vapor displaces an equal volume of air, the new density is calculated as
$$\rho_{new} = \left(1-\frac{29-18}{29}\frac{m}{M}\right)\rho_{old}$$
where 18 is the molecular mass of water, and 29 is the molecular mass of air.
The temperature drops by $$T_{new} = \left(1-\frac{2300}{T_{old}}\frac{m}{M}\right)T_{old}$$
Since the density of the air scales with the inverse temperature (universal gas law) we can find the actual change in density by dividing the new density by the new temperature to see whether the change in average molecular mass or the cooling is the more important factor in determining the direction of the air flow.
If we assume that $\frac{m}{M}$ is small, the fraction $\frac{\rho}{T}$ becomes
$$\frac{\rho_{new}}{ T_{new}} \approx \left(1 - \left(\frac{11}{29}-\frac{2300}{300}\right)\right)\frac{\rho_{old}}{T_{old}}$$
It is clear that the cooling term is much greater than the density term - so the air, being cooled by the leaves of the tree, will drift down and cool the air under the tree more.
It seems to me that this will push the balance of temperature towards a small, but measurable, drop in air temperature under the tree - even if you use a measuring instrument that is insensitive to radiation.
