Riesz's lemma (after Frigyes Riesz) is a lemma in functional analysis. It specifies (often easy to check) conditions that guarantee that a subspace in a normed vector space is dense. The lemma may also be called the Riesz lemma or Riesz inequality. It can be seen as a substitute for orthogonality when the normed space is not an inner product space.
Riesz's lemma^{[1]} — Let be a closed proper vector subspace of a normed space and let be any real number satisfying Then there exists a vector in of unit norm such that for all in
If is a reflexive Banach space then this conclusion is also true when ^{[2]}
Metric reformulation
As usual, let denote the canonical metric induced by the norm, call the set of all vectors that are a distance of from the origin the unit sphere, and denote the distance from a point to the set by
Using this new notation, the conclusion of Riesz's lemma may be restated more succinctly as: holds for some
Using this new terminology, Riesz's lemma may also be restated in plain English as:
The proof^{[3]} can be found in functional analysis texts such as Kreyszig.^{[4]} An online proof from Prof. Paul Garrett is available.
Consider any and denote its distance from by . Clearly, since is closed. Take any . By the definition of an infimum, there is a such that

(1)

(note that since ). Let
Then , and we show that for every . We have
where
The form of shows that . Hence , by the definition of . Writing out and using (1), we obtain
Since was arbitrary, this completes the proof.
Minimum distances not satisfying the hypotheses
When is trivial then it has no proper vector subspace and so Riesz's lemma holds vacuously for all real numbers The remainder of this section will assume that which guarantees that a unit vector exists.
The inclusion of the hypotheses can be explained by considering the three cases: , and The lemma holds when since every unit vector satisfies the conclusion The hypotheses is included solely to exclude this trivial case and is sometimes omitted from the lemma's statement.
Riesz's lemma is always false when because for every unit vector the required inequality fails to hold for (since ). Another of consequence of being impossible is that the inequality holds if and only if equality holds.
This leaves only the case for consideration, in which case the statement of Riesz's lemma becomes:
When is a Banach space, then this statement is true if and only if is a reflexive space.^{[2]} Explicitly, a Banach space is reflexive if and only if for every closed proper vector subspace there is some vector on the unit sphere of that is always at least a distance of away from the subspace.
For example, if the reflexive Banach space is endowed with the usual Euclidean norm and if is the plane then the points satisfy the conclusion If is axis then every point belonging to the unit circle in the plane satisfies the conclusion But if was endowed with the taxicab norm (instead of the Euclidean norm), then the conclusion would be satisfied by every point belonging to the "diamond" in the plane (a square with vertices at and )
In a nonreflexive Banach space, such as the Lebesgue space of all bounded sequences, Riesz's lemma does not hold for ^{[5]}
However, every finite dimensional normed space is a reflexive Banach space, so Riesz's lemma does holds for when the normed space is finitedimensional, as will now be shown. When the dimension of is finite then the closed unit ball is compact. Since the distance function is continuous, its image on the closed unit ball must be a compact subset of the real line, proving the claim.
Riesz's lemma guarantees that for any given every infinitedimensional normed space contains a sequence of (distinct) unit vectors satisfying for or stated in plain English, these vectors are all separated from each other by a distance of more than while simultaneously also all lying on the unit sphere. Such an infinite sequence of vectors cannot be found in the unit sphere of any finite dimensional normed space (just consider for example the unit circle in ).
This sequence can be constructed by induction for any constant Start by picking any element from the unit sphere. Let be the linear span of and (using Riesz's lemma) pick from the unit sphere such that
This sequence contains no convergent subsequence, which implies that the closed unit ball is not compact.
Riesz's lemma can be applied directly to show that the unit ball of an infinitedimensional normed space is never compact. This can be used to characterize finite dimensional normed spaces: if is a normed vector space, then is finite dimensional if and only if the closed unit ball in is compact.
More generally, if a topological vector space is locally compact, then it is finite dimensional. The converse of this is also true. Namely, if a topological vector space is finite dimensional, it is locally compact.^{[6]} Therefore local compactness characterizes finitedimensionality. This classical result is also attributed to Riesz. A short proof can be sketched as follows: let be a compact neighborhood of the origin in By compactness, there are such that
We claim that the finite dimensional subspace spanned by is dense in or equivalently, its closure is Since is the union of scalar multiples of it is sufficient to show that By induction, for every
The spectral properties of compact operators acting on a Banach space are similar to those of matrices. Riesz's lemma is essential in establishing this fact.
As detailed in the article on infinitedimensional Lebesgue measure, this is useful in showing the nonexistence of certain measures on infinitedimensional Banach spaces. Riesz's lemma also shows that the identity operator on a Banach space is compact if and only if is finitedimensional.^{[8]}