In finite-dimensional [normed vector spaces](/page/Normed%20Vector%20Space), the geometry of the unit ball is tightly constrained by compactness. The closed unit ball $\overline{B}(0, 1) = \{x \in X : \|x\| \le 1\}$ in $\mathbb{R}^n$ is compact by the Heine-Borel theorem, and this compactness guarantees that many optimisation problems — finding a nearest point in a closed subspace, attaining the norm of a functional, extracting convergent subsequences from bounded sequences — have solutions. A recurring ingredient in these arguments is the ability to find unit vectors that are "well separated" from a given subspace: if $Y \subsetneq \mathbb{R}^n$ is a proper closed subspace, then the distance function $x \mapsto \operatorname{dist}(x, Y)$ is continuous on the compact set $\{x \in \mathbb{R}^n : \|x\| = 1\}$, and since it is strictly positive there (because $Y$ is closed and does not contain every unit vector), it attains its maximum. In particular, there exists a unit vector $x_0$ with $\operatorname{dist}(x_0, Y) = 1$ — perfect separation.

In infinite-dimensional spaces, the closed unit ball is never compact, and this argument breaks down. The distance function is still continuous, and it is still strictly positive on the unit sphere, but the supremum need not be attained. Can we still find unit vectors that are *almost* maximally separated from a closed subspace — not at distance $1$, but at distance arbitrarily close to $1$? Riesz's Lemma, proved by Frigyes Riesz in 1918, answers this question affirmatively. It is a simple geometric result, but its consequences are far-reaching: it provides the fundamental obstruction to compactness in infinite dimensions, it is the engine behind the proof that a [normed vector space](/page/Normed%20Vector%20Space) is finite-dimensional if and only if its closed unit ball is [compact](/page/Compact%20Space), and it gives the basic tool for constructing well-separated sequences in infinite-dimensional spaces.

[example: The Distance-One Problem in $\ell^2$] Consider $X = \ell^2(\mathbb{N})$ and the closed subspace $Y = \{x \in \ell^2 : x_1 = 0\}$, which consists of all sequences whose first coordinate vanishes. For any $x = (x_1, x_2, \ldots) \in \ell^2$, the nearest point in $Y$ is the projection $P_Y x = (0, x_2, x_3, \ldots)$, and \begin{align*} \operatorname{dist}(x, Y) = \|x - P_Y x\|_{\ell^2} = |x_1|. \end{align*} On the unit sphere $\{x \in \ell^2 : \|x\|_{\ell^2} = 1\}$, the distance to $Y$ equals $|x_1|$, which is maximised at $x_0 = e_1 = (1, 0, 0, \ldots)$, giving $\operatorname{dist}(e_1, Y) = 1$. So in this case, a unit vector at distance exactly $1$ from $Y$ exists. This is not a coincidence — $\ell^2$ is a [Hilbert space](/page/Hilbert%20Space), and the orthogonal projection theorem guarantees that the distance to any closed subspace is attained. The same is true in every Hilbert space: given a proper closed subspace $Y$, any unit vector in the orthogonal complement $Y^\perp$ satisfies $\operatorname{dist}(x, Y) = 1$. [/example]

The Hilbert space example might suggest that distance $1$ is always achievable. But the orthogonal projection theorem depends on the inner product structure, not just the norm. In general [Banach spaces](/page/Banach%20Space), no orthogonal complement exists, and the geometry of the unit ball can be so skewed that no unit vector achieves distance exactly $1$ from a given closed subspace — the supremum of $\operatorname{dist}(\cdot, Y)$ on the unit sphere is $1$, but it is not attained. Riesz's Lemma guarantees that the supremum can be approached as closely as desired, even when it cannot be reached.

## Definition

The lemma addresses a fundamental geometric question about [normed vector spaces](/page/Normed%20Vector%20Space): given a proper closed subspace, how close to the subspace must unit vectors be? The answer is that no unit vector is forced to be close — for any tolerance $\theta < 1$, one can always find a unit vector at distance at least $\theta$ from the subspace.

[quotetheorem:1222]

The construction of $x_\theta$ is explicit and elementary. Choose any $z \in X \setminus Y$. Since $Y$ is closed, $d := \operatorname{dist}(z, Y) > 0$. By the definition of infimum, there exists $y_0 \in Y$ with

\begin{align*} \|z - y_0\| < \frac{d}{\theta}. \end{align*}

Set $x_\theta := (z - y_0)/\|z - y_0\|$. Then $\|x_\theta\| = 1$, and for any $y \in Y$:

\begin{align*} \|x_\theta - y\| &= \left\|\frac{z - y_0}{\|z - y_0\|} - y\right\| = \frac{1}{\|z - y_0\|}\|z - y_0 - \|z - y_0\| y\| \\ &= \frac{1}{\|z - y_0\|}\|z - \underbrace{(y_0 + \|z - y_0\| y)}_{\in Y}\| \\ &\ge \frac{d}{\|z - y_0\|} > \frac{d}{d/\theta} = \theta. \end{align*}

The element $y_0 + \|z - y_0\| y$ belongs to $Y$ because $Y$ is a subspace (closed under addition and scalar multiplication). Since $y \in Y$ was arbitrary, $\operatorname{dist}(x_\theta, Y) \ge \theta$.

[remark: Role of Closedness] The hypothesis that $Y$ is closed is essential. If $Y$ is merely a proper subspace (not closed), then $\operatorname{dist}(z, Y)$ can be zero for $z \notin Y$ — specifically, $z$ could lie in $\overline{Y} \setminus Y$. In this case, the quantity $d = \operatorname{dist}(z, Y) = 0$, and the construction collapses. For instance, in $X = C([0,1])$ with the supremum norm, let $Y$ be the subspace of polynomials. Then $Y$ is dense in $X$ (by the [Weierstrass approximation theorem](/page/Weierstrass%20Approximation%20Theorem)), so $Y$ is not closed and $\operatorname{dist}(f, Y) = 0$ for every $f \in X$, even though $Y \neq X$. No conclusion of the Riesz Lemma type is possible: since $Y$ is dense, every unit vector is arbitrarily close to $Y$. [/remark]

## The Failure at $\theta = 1$

The restriction $\theta \in (0, 1)$ in Riesz's Lemma — excluding $\theta = 1$ — is the most delicate aspect of the result. It is natural to ask: can one always find a unit vector $x$ with $\operatorname{dist}(x, Y) = 1$, achieving the maximum possible distance? The answer depends on the geometry of the space, and understanding exactly when and why $\theta = 1$ fails reveals a deep connection to [reflexivity](/page/Reflexive%20Space).

### Spaces Where $\theta = 1$ Succeeds

In two important classes of spaces, the distance $1$ is always attained.

**Finite-dimensional spaces.** If $\dim X < \infty$, the unit sphere $S := \{x \in X : \|x\| = 1\}$ is compact, and the continuous function $x \mapsto \operatorname{dist}(x, Y)$ attains its supremum on $S$. Since $Y \subsetneq X$ and $Y$ is closed, this supremum is positive (any $z \in X \setminus Y$ gives a positive value when normalised) and at most $1$ (because $0 \in Y$ and $\|x\| = 1$ for $x \in S$). A direct argument shows the supremum equals $1$: for any $x \in S$ and $y \in Y$, the triangle inequality gives $\|x - y\| \ge \|x\| - \|y\| = 1 - \|y\|$, and in particular $\operatorname{dist}(x, Y) \le \|x\| = 1$. The supremum is exactly $1$, and compactness of $S$ guarantees it is attained.

**Hilbert spaces.** If $X$ is a [Hilbert space](/page/Hilbert%20Space) and $Y$ is a proper closed subspace, the orthogonal decomposition $X = Y \oplus Y^\perp$ with $Y^\perp \neq \{0\}$ provides a unit vector $x \in Y^\perp$ with

\begin{align*} \operatorname{dist}(x, Y) = \inf_{y \in Y} \|x - y\| = \|x\| = 1, \end{align*}