This article needs additional citations for verification. (November 2012) |

**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 one is not in an inner product space.

## The result

Riesz's Lemma.LetXbe a normed space,Ybe a closed proper subspace ofXand α be a real number with 0 < α < 1. Then there exists anxinXwith |x| = 1 such that |x−y| ≥ α for allyinY.^{[1]}

*Remark 1.* For the finite-dimensional case, equality can be achieved. In other words, there exists *x* of unit norm such that *d*(*x*, *Y*) = 1. When dimension of *X* is finite, the unit ball *B* ⊂ *X* is compact. Also, the distance function *d*(· , *Y*) is continuous. Therefore its image on the unit ball *B* must be a compact subset of the real line, proving the claim.

*Remark 2.* The space ℓ_{∞} of all bounded sequences shows that the lemma does not hold for α = 1.

The proof can be found in functional analysis texts such as Kreyszig. An online proof from Prof. Paul Garrett is available.

## Some consequences

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.

Riesz's lemma guarantees that any infinite-dimensional normed space contains a sequence of unit vectors {*x _{n}*} with for 0 <

*α*< 1. This is useful in showing the non-existence of certain measures on infinite-dimensional Banach spaces. Riesz's lemma also shows that the identity operator on a Banach space

*X*is compact if and only if

*X*is finite-dimensional.

^{[2]}

One can also use this lemma to characterize finite dimensional normed spaces: if X is a normed vector space, then X is finite dimensional if and only if the closed unit ball in X is compact.

### Characterization of finite dimension

Riesz's lemma can be applied directly to show that the unit ball of an infinite-dimensional normed space *X* is never compact: Take an element *x*_{1} from the unit sphere. Pick *x _{n}* from the unit sphere such that

- for a constant 0 <
*α*< 1, where*Y*_{n−1}is the linear span of {*x*_{1}...*x*_{n−1}} and .

Clearly {*x*_{n}} contains no convergent subsequence and the noncompactness of the unit ball follows.

More generally, if a topological vector space *X* 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.^{[3]} Therefore local compactness characterizes finite-dimensionality. This classical result is also attributed to Riesz. A short proof can be sketched as follows: let *C* be a compact neighborhood of 0 ∈ *X*. By compactness, there are *c*_{1}, ..., *c _{n}* ∈

*C*such that

We claim that the finite dimensional subspace *Y* spanned by {*c _{i}*} is dense in

*X*, or equivalently, its closure is

*X*. Since

*X*is the union of scalar multiples of

*C*, it is sufficient to show that

*C*⊂

*Y*. Now, by induction,

for every *m*. But compact sets are bounded, so *C* lies in the closure of *Y*. This proves the result. For a different proof based on Hahn-Banach Theorem see.^{[4]}

## See also

## References

**^**Rynne, Bryan P.; Youngson, Martin A. (2008).*Linear Functional Analysis*(2nd ed.). London: Springer. p. 47. ISBN 978-1848000049.**^**Kreyszig (1978, Theorem 2.5-3, 2.5-5)**^**https://terrytao.wordpress.com/2011/05/24/locally-compact-topological-vector-spaces/**^**https://www.emis.de/journals/PM/51f2/pm51f205.pdf/

- Kreyszig, Erwin (1978),
*Introductory functional analysis with applications*, John Wiley & Sons, ISBN 0-471-50731-8