This is a rant on Riesz’s lemma.
Riesz’s lemma- Let there be a vector space and a closed proper subspace . Then , there exists a such that , where , and .
A proof is commonly available. What we will discuss here is the thought behind the proof.
For any random and , write . Let . Then . Also, there exists a such that . Then . Because the vector space is closed under scalar multiplication, we have effectively proved for any and .
If there is some other vector such that , then .
Hence, one part of Riesz’s lemma, that of exceeding is satisfied by every vector . The thoughts to take away from this is dividing by or a number less than increases everything, even a small increase from the infimum exceeds terms of a sequence converging to the infimum, and every arbitrary term in the sequence is greater than the infimum. When we say can be any number in the interval , we know we’re skirting with boundaries. We could aso have thought of a proof in this direction: let . Then . However, for an arbitrary , .
Hence, for every , .
Now what about ? This condition is satisfied only when in the expression .
Hence, over in all, for every vector , there are infinite vectors which satisfy the condition of Riesz’s lemma. Also, for every such , there is AT LEAST one unit vector which satisfies Riesz’s lemma (there can be more than one). Hence, to think there can be only one unit vector in which satisfies Riesz’s lemma would be erroneous.