Blueshift and increase in energy! 
Imagine that there is a car and it is not moving but its headlights are on. There is a wall in front of the car but is very far away. Right now energy is being used only in switching on the headlights. Now the car starts moving at a very high speed.

As I have shown in the picture, there is a blueshift of light and so the energy of light emitted per unit time has increased. Now my question is that from where does this extra energy come from.
Some arguments that prove that extra energy is generated. If the car was moving without the headlights off but at the same speed, the energy would have been used in the movement of the car. Now if the car was not moving and only the headlight was on, the energy would have been used in powering the headlight. But when we take both the cases simultaneously, then we see that there is an increase in the net energy. For further explanation I will give some equations.
Case 1 when the car is moving but the headlights are off
$Q_1 = \frac{dE}{dt} = \frac{d\sqrt{p^2c^2 + m^2c^4}}{dt} = 0;$
Case 2 when the car is not moving but the headlights are on
$Q_2 = \frac{dE}{dt} = \frac{d(mc^2)}{dt} < 0$ since energy is being radiated by the lights on;
Case 3 when the car is moving and the headlights are on
$Q_3 = \frac{dE}{dt} = \frac{d\sqrt{p^2c^2 + m^2c^4}}{dt} < Q_2$, because the power of radiation is higher than in Case 2. This is because the light is blue-shifted and its quanta have higher energy.
So from where does this extra energy of light come from?
 A: The extra energy comes from the kinetic energy of the moving car. The radiated light is carrying away some momentum and is decreasing the speed (and therefore kinetic energy) of the car.
A: There is no extra energy - the blueshift is due to an increase in the observed frequency, not the actual frequency.  The light is still being emitted at the same frequency so the same amount of energy is used.  
A: The reason for this is that energy is not the same for different observers. Conservation of energy makes it sound like a person in the car and a person standing on the road should measure the same energy in the system, but they won't. It just means that when some event happens, the energy in the system will be the same before and after, and the two observers will agree on that, but they can be talking about different amounts of energy in their reference frame. A simple example: a person on the road will say the car has an immense kinetic energy, but a person in the car will be at rest relative to the car so they'll think the car has no kinetic energy. The difference in light frequencies is like that: two observers see different colored light, but they will still agree that when something happens, whatever energy they measure will remain constant in their view of the system.
A: The short version is this: the observer and car disagree about the amount of energy that the car has. They agree on how much the relative velocity between them is changing, but disagree about the amount of 'braking energy' which is captured by slowing the car down. That captured energy goes into the photons released. Different reference frames do not generally agree on the total energy within a system. This is because different reference frames disagree on the 'direction' of time vs. space. What one frame sees as energy, another sees as a more rapid advancement in proper time advancing for the observed object. Don't worry if you don't understand this last part.
Both observers agree on the rest mass of the car after any given photon emission. Both agree on the number of photons emitted and exactly at what point in spacetime that emission occurred. Both agree on their relative velocity, both initially and after each individual photon is emitted.
However, there is an asymmetry in this thought experiment, which is which of the two observers is being accelerated. Each time a photon is emitted, the car is being decelerated. This is crucial. There are two ways to reduce their relative velocity-- either the observer changes velocity, or the car does. In this experiment, the car is changing velocity because it is the one emitting photons.
When an object is moving quickly in a reference frame (comparable with the speed of light), it takes a large amount of momentum ($p = \gamma m v; \gamma > 1$) to slow them down, and a large reduction in the car's kinetic energy actually corresponds to only a small change in relative velocity:
$$
\mathrm d E = m c^2 \gamma \beta \; \mathrm d\beta\; , \quad \gamma=\left( 1 - \beta^2\right)^{-1/2} \; , \quad \gamma > 1
$$
where $\beta = v/c$. Or if you prefer:
$$
\mathrm d E = \gamma m v \, \mathrm d v \, \quad \text{where} \quad \gamma >1 .
$$
According to the observer, the car's velocity changes only a little, but a lot of 'braking energy' went into producing the photon.
When an object is slow or static in a reference frame (the car's perspective; $\gamma \approx 1$; $p \approx m v$), there is only a small 'braking energy' associated with that same decrease in velocity:
$$
dE \approx m c^2 \beta \; \mathrm d \beta = m v \, \mathrm d v \,.
$$
This is the same expression as above, except $\gamma \approx 1$ rather than $\gamma > 1$. From the car's point of view, there is less energy loss associated with a small deceleration-- a smaller photon energy.
While both observer's agree with the amount of kinetic velocity that the car is losing, they disagree about the amount of energy required to slow it down.
