In particular, the mapping pH from E into H is continuous.
Let x,y be two points of E. Put a=pH(x)−x,b=pH(y)−pH(x),c=y−pH(y). By formula (15) (V, p. 10) we have R⟨a∣b⟩≥0 and R⟨c∣b⟩≥0. We also have a+b+c=y−x, which gives,
∥x−y∥2=∥a+b+c∥2=∥b∥2+∥a+c∥2+2R⟨a∣b⟩+2R⟨c∣b⟩
≥∥b∥2=∥pH(x)−pH(y)∥2.
This proves corollary 2.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.