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*} where the second equality uses the Pythagorean identity: $\|x - y\|^2 = \|x\|^2 + \|y\|^2 \ge \|x\|^2$ for $x \in Y^\perp$ and $y \in Y$. ### A Banach Space Where $\theta = 1$ Fails The failure of $\theta = 1$ is a genuine phenomenon in Banach spaces that lack sufficient geometric structure. The mechanism is the following: the question of whether a unit vector at distance exactly $1$ from a closed hyperplane $Y = \ker \Lambda$ exists reduces to whether the bounded linear functional $\Lambda \in X^*$ attains its norm on the unit ball. If $\Lambda$ does not attain its norm, then no unit vector achieves distance $1$ from $\ker \Lambda$. To see this connection, recall that for any $f \in X$ and any hyperplane $Y = \ker \Lambda$ (where $\Lambda \in X^*$ is nonzero): \begin{align*} \operatorname{dist}(f, Y) = \frac{|\Lambda(f)|}{\|\Lambda\|_{X^*}}. \end{align*} For a unit vector $f$ with $\|f\| = 1$, we have $|\Lambda(f)| \le \|\Lambda\|_{X^*}$, so $\operatorname{dist}(f, Y) \le 1$. Equality $\operatorname{dist}(f, Y) = 1$ holds if and only if $|\Lambda(f)| = \|\Lambda\|_{X^*}$ with $\|f\| = 1$ — that is, if and only if $\Lambda$ attains its norm at $f$. [example: Non-Attainment of Distance One in $C([0,1])$] Let $X = C([0,1])$ with the supremum norm $\|f\|_\infty = \sup_{t \in [0,1]} |f(t)|$. Define the bounded linear functional \begin{align*} \Lambda: C([0,1]) &\to \mathbb{R} \\ f &\mapsto \int_0^1 (2t - 1) f(t) \, d\mathcal{L}^1(t). \end{align*} The operator norm of $\Lambda$ is \begin{align*} \|\Lambda\|_{X^*} = \sup_{\|f\|_\infty \le 1} \left|\int_0^1 (2t-1) f(t) \, d\mathcal{L}^1(t)\right| = \int_0^1 |2t - 1| \, d\mathcal{L}^1(t). \end{align*} The last equality is a standard fact about $L^1$-functionals on $C([0,1])$: the supremum is achieved in the limit by functions that approximate $\operatorname{sgn}(2t - 1)$. Computing the integral: \begin{align*} \int_0^1 |2t - 1| \, d\mathcal{L}^1(t) &= \int_0^{1/2} (1 - 2t) \, d\mathcal{L}^1(t) + \int_{1/2}^1 (2t - 1) \, d\mathcal{L}^1(t) \\ &= \left[t - t^2\right]_0^{1/2} + \left[t^2 - t\right]_{1/2}^1 \\ &= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}. \end{align*} We claim that $\Lambda$ does **not** attain its norm. Suppose for contradiction that some $f \in C([0,1])$ with $\|f\|_\infty = 1$ satisfies $|\Lambda(f)| = 1/2$. Without loss of generality, assume $\Lambda(f) = 1/2$ (replacing $f$ by $-f$ if necessary). Then \begin{align*} \frac{1}{2} = \int_0^1 (2t-1) f(t) \, d\mathcal{L}^1(t) \le \int_0^1 |2t - 1| \cdot |f(t)| \, d\mathcal{L}^1(t) \le \int_0^1 |2t-1| \, d\mathcal{L}^1(t) = \frac{1}{2}. \end{align*} Equality in the first inequality requires $(2t-1)f(t) = |2t-1| \cdot |f(t)|$ for $\mathcal{L}^1$-almost every $t$. This forces $f(t) \ge 0$ for $t > 1/2$ and $f(t) \le 0$ for $t < 1/2$. Equality in the second inequality requires $|f(t)| = 1$ for $\mathcal{L}^1$-almost every $t$ where $|2t - 1| > 0$ — that is, for $\mathcal{L}^1$-almost every $t \neq 1/2$. Since $f$ is continuous, this forces $|f(t)| = 1$ for all $t \in [0,1]$, which combined with the sign condition gives $f(t) = +1$ for $t > 1/2$ and $f(t) = -1$ for $t < 1/2$. But a continuous function cannot jump from $-1$ to $+1$ at $t = 1/2$. The intermediate value theorem guarantees that $f$ must pass through $0$ at some point near $t = 1/2$, contradicting $|f(t)| = 1$ everywhere. Therefore $\Lambda$ does not attain its norm. Setting $Y := \ker \Lambda$, the identity $\operatorname{dist}(f, Y) = |\Lambda(f)|/\|\Lambda\|_{X^*}$ shows that $\operatorname{dist}(f, Y) < 1$ for every unit vector $f \in X$. The supremum $\sup_{\|f\|_\infty = 1} \operatorname{dist}(f, Y) = 1$ is approached by the sequence $f_k(t) := \min(1, \max(-1, k(2t-1)))$ of smoothed sign functions, each with $\|f_k\|_\infty = 1$ and $\Lambda(f_k) \to 1/2$, but the supremum is never attained. [/example] [explanation: The Connection to Reflexivity] The failure of $\theta = 1$ is not an isolated curiosity — it is governed by a deep structural property of the Banach space. James's theorem (1957) provides a precise characterisation: a [Banach space](/page/Banach%20Space) is [reflexive](/page/Reflexive%20Space) if and only if every continuous linear functional attains its norm on the closed unit ball. Combining this with the distance formula $\operatorname{dist}(f, \ker \Lambda) = |\Lambda(f)|/\|\Lambda\|_{X^*}$ yields: - In every **reflexive** Banach space, $\theta = 1$ is achievable for every proper closed hyperplane (since every functional attains its norm), and by induction on codimension, for every proper closed subspace of finite codimension. - In every **non-reflexive** Banach space, there exists at least one continuous linear functional that does not attain its norm, and therefore at least one closed hyperplane for which $\theta = 1$ fails. The space $C([0,1])$ in the example above is not reflexive (its dual is the space of signed Radon measures on $[0,1]$, whose dual is strictly larger than $C([0,1])$), consistent with the existence of non-norm-attaining functionals. This reveals a structural hierarchy. Riesz's Lemma with $\theta < 1$ holds in all normed spaces — no geometric hypothesis is needed. The upgrade to $\theta = 1$ requires either finite-dimensionality (where compactness of the unit sphere forces attainment), or the Hilbert space structure (where orthogonal complements exist), or more generally, reflexivity (where James's theorem guarantees norm attainment). The gap between $\theta < 1$ and $\theta = 1$ is precisely the gap between general normed spaces and reflexive Banach spaces. [/explanation] ## Non-Compactness of the Unit Ball The most celebrated application of Riesz's Lemma is the proof that the closed unit ball of an infinite-dimensional normed space is never [compact](/page/Compact%20Space). This result — sometimes called the **Riesz theorem on finite-dimensional spaces** — provides the fundamental structural dividing line between finite- and infinite-dimensional analysis. The difficulty this result addresses is the following: in finite dimensions, compactness of bounded closed sets (Heine-Borel) is the workhorse of existence theory. One extracts convergent subsequences from bounded sequences, attains infima and suprema of continuous functions, and guarantees that continuous images of compact sets are compact. If these tools are unavailable in infinite dimensions, then entirely new methods — [weak topologies](/page/Weak%20Topology), the [Banach-Alaoglu theorem](/page/Banach-Alaoglu%20Theorem), compactness criteria like Arzela-Ascoli and Rellich-Kondrachov — are needed. Riesz's Lemma explains *why* these new methods are necessary by exposing the precise mechanism of failure. [quotetheorem:878] The forward direction (finite dimension implies compactness) follows from the Heine-Borel theorem: in $\mathbb{R}^n$, every closed and bounded set is compact, and any finite-dimensional normed space is linearly homeomorphic to $\mathbb{R}^n$ with some norm (all norms on $\mathbb{R}^n$ are equivalent). The reverse direction (compactness implies finite dimension) is where Riesz's Lemma enters. Suppose $\dim X = \infty$. We construct a sequence $(x_k)_{k=1}^\infty$ of unit vectors with $\|x_j - x_k\| \ge 1/2$ for all $j \neq k$. Such a sequence has no Cauchy subsequence, hence no convergent subsequence, so $\overline{B}(0, 1)$ is not sequentially compact, hence not compact. The construction proceeds inductively. Pick any unit vector $x_1 \in X$. The span $Y_1 := \operatorname{span}\{x_1\}$ is a finite-dimensional (hence closed) proper subspace of $X$. By Riesz's Lemma with $\theta = 1/2$, there exists $x_2 \in X$ with $\|x_2\| = 1$ and $\operatorname{dist}(x_2, Y_1) \ge 1/2$. In particular, $\|x_2 - x_1\| \ge 1/2$. Now let $Y_2 := \operatorname{span}\{x_1, x_2\}$, which is again closed (finite-dimensional) and proper (since $\dim X = \infty$). By Riesz's Lemma with $\theta = 1/2$, there exists $x_3$ with $\|x_3\| = 1$ and $\operatorname{dist}(x_3, Y_2) \ge 1/2$. Since $x_1, x_2 \in Y_2$, this gives $\|x_3 - x_1\| \ge 1/2$ and $\|x_3 - x_2\| \ge 1/2$. Continuing inductively, at step $k$ we set $Y_{k-1} := \operatorname{span}\{x_1, \ldots, x_{k-1}\}$ and apply Riesz's Lemma to obtain $x_k$ with $\|x_k\| = 1$ and $\|x_k - x_j\| \ge \operatorname{dist}(x_k, Y_{k-1}) \ge 1/2$ for all $j < k$. The sequence $(x_k)_{k=1}^\infty$ lies in $\overline{B}(0, 1)$ and satisfies $\|x_j - x_k\| \ge 1/2$ for all $j \neq k$. [explanation: Why the Proof Requires Riesz's Lemma, Not Just Linear Independence] A naive approach to showing non-compactness might proceed as follows: since $X$ is infinite-dimensional, choose a linearly independent sequence $(v_k)_{k=1}^\infty$ and normalise to get unit vectors $x_k = v_k/\|v_k\|$. The problem is that linear independence does not control distances: the unit vectors $x_k$ might converge to each other, or even be Cauchy, despite being linearly independent. For a concrete illustration, consider $\ell^2(\mathbb{N})$ and the sequence $x_k = \sqrt{1 - 1/k^2} \, e_1 + (1/k) \, e_k$ for $k \ge 1$. Each $x_k$ satisfies $\|x_k\|_{\ell^2} = 1$, and the vectors $\{x_k\}$ are linearly independent (since $x_k$ is the only element involving $e_k$). However, \begin{align*} \|x_j - x_k\|_{\ell^2}^2 &= \left(\sqrt{1 - 1/j^2} - \sqrt{1 - 1/k^2}\right)^2 + \frac{1}{j^2} + \frac{1}{k^2} \to 0 \quad \text{as } j, k \to \infty. \end{align*} The sequence converges to $e_1$, and in particular it is Cauchy. Linear independence gives no protection against this clustering. Riesz's Lemma provides what linear independence alone cannot: a *quantitative* separation guarantee. By choosing $x_k$ to be at distance $\ge 1/2$ from the span of all previous vectors, one ensures that the sequence is uniformly separated. This is the critical step that makes the non-compactness argument work. [/explanation] [remark: The Role of $\theta = 1/2$] The choice $\theta = 1/2$ in the construction is convenient but not essential. Any fixed $\theta \in (0, 1)$ produces a sequence of unit vectors with mutual distances at least $\theta$, which suffices to prevent Cauchy subsequences. The choice $\theta = 1/2$ gives a clean separation constant that is easy to work with. One cannot take $\theta = 1$ in general, because as we have seen, $\theta = 1$ can fail in non-reflexive spaces — and the argument must work in *every* normed space. [/remark] ## The Riesz Characterisation of Finite-Dimensional Spaces The non-compactness result above is half of a fundamental characterisation. The full statement gives equivalent conditions for finite-dimensionality in terms of compactness and [total boundedness](/page/Total%20Boundedness), unifying several perspectives on the finite-dimensional/infinite-dimensional divide. [quotetheorem:1224] The implications $(1) \Rightarrow (4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (5) \Rightarrow (1)$ form a complete cycle. The direction $(1) \Rightarrow (4)$ is the Heine-Borel theorem: every bounded set in a finite-dimensional normed space lies in a closed ball, which is compact, hence totally bounded. The direction $(3) \Rightarrow (2)$ uses the fact that a finite-dimensional normed space is complete (all norms are equivalent to the Euclidean norm, and $\mathbb{R}^n$ is complete), so the closed unit ball is a complete and totally bounded metric space, hence compact. The direction $(5) \Rightarrow (1)$ is the contrapositive of the Riesz Lemma construction: if $\dim X = \infty$, the $1/2$-separated sequence constructed above is bounded with no convergent subsequence. The equivalence of (2) and (3) for the closed unit ball deserves attention. In a general metric space, total boundedness is strictly weaker than compactness — one also needs completeness. But for the closed unit ball of a normed space, total boundedness alone forces finite-dimensionality (via the failure of (3) when $\dim X = \infty$), which in turn gives compactness by Heine-Borel. The key observation is that the Riesz Lemma construction provides a sequence of unit vectors with $\|x_j - x_k\| \ge 1/2$, so the unit ball cannot be covered by finitely many balls of radius $1/4$ — that is, the unit ball is not totally bounded. Total boundedness and compactness are thus equivalent *for unit balls of normed spaces*, even without assuming completeness. ## Compact Operators and Riesz's Lemma The structural dividing line established by Riesz's Lemma — compact unit ball if and only if finite-dimensional — has a natural operator-theoretic counterpart. [definition: Compact Operator] A [bounded linear operator](/page/Bounded%20Linear%20Operator) $T \in \mathcal{L}(X, Y)$ between normed spaces is called **compact** if $T(\overline{B}(0,1))$ has compact closure in $Y$, or equivalently, if every bounded sequence $(x_k)$ in $X$ has a subsequence $(x_{k_j})$ such that $(Tx_{k_j})$ converges in $Y$. [/definition] Riesz's characterisation tells us immediately that the identity operator $\operatorname{Id}: X \to X$ is compact if and only if $\dim X < \infty$. More broadly, Riesz's Lemma provides the standard tool for proving that specific operators are *not* compact. [example: The Identity Is Not Compact on Infinite-Dimensional Spaces] Let $X$ be any infinite-dimensional normed space. The Riesz Lemma construction produces a sequence $(x_k)_{k=1}^\infty$ in the closed unit ball with $\|x_j - x_k\| \ge 1/2$ for $j \neq k$. Since $\operatorname{Id}(x_k) = x_k$, this sequence has no convergent subsequence under the identity, so $\operatorname{Id}_X$ is not compact. For a concrete instance, take $X = \ell^2(\mathbb{N})$ and the standard basis $(e_k)_{k=1}^\infty$. Each $e_k$ lies in the unit ball, and $\|e_j - e_k\|_{\ell^2} = \sqrt{2}$ for $j \neq k$, which is even stronger than the $1/2$-separation that Riesz's Lemma guarantees in general. [/example] The non-compactness of the identity is the foundation for the spectral theory of compact operators. Since the identity on an infinite-dimensional space is not compact, any compact operator $T$ on such a space must "compress" the unit ball — the image $T(\overline{B}(0,1))$ is a strictly smaller set (in the sense of total boundedness) than $\overline{B}(0,1)$ itself. This compression is what gives compact operators their finite-dimensional-like spectral behaviour. [example: Non-Compactness of Critical Sobolev Embeddings] The Riesz Lemma philosophy — constructing separated sequences to disprove compactness — extends to function space embeddings. Consider the Sobolev embedding $W^{1,p}(U) \hookrightarrow L^{p^*}(U)$ at the critical exponent $p^* = np/(n-p)$ for $1 \le p < n$ and bounded open $U \subset \mathbb{R}^n$. This embedding is continuous (the Gagliardo-Nirenberg-Sobolev inequality) but **not** compact. To exhibit non-compactness, fix a smooth function $\phi \in C_c^\infty(\mathbb{R}^n)$ with $\|\phi\|_{L^{p^*}} = 1$ and $\operatorname{supp} \phi \subset B(0,1)$, and define the rescaled sequence \begin{align*} \phi_k(x) := k^{n/p^*} \phi(k(x - x_0)) \end{align*} for a fixed $x_0 \in U$ and $k$ large enough that $\operatorname{supp} \phi_k \subset U$. A change of variables $y = k(x - x_0)$ gives \begin{align*} \|\phi_k\|_{L^{p^*}(U)}^{p^*} = k^n \int_{B(0,1)} |\phi(y)|^{p^*} k^{-n} \, d\mathcal{L}^n(y) = \|\phi\|_{L^{p^*}}^{p^*} = 1, \end{align*} and a similar computation shows $\|\nabla \phi_k\|_{L^p(U)}$ remains bounded (by the scale-invariance built into the critical exponent $p^*$). The sequence $(\phi_k)$ is bounded in $W^{1,p}(U)$. However, the supports $\operatorname{supp} \phi_k$ shrink to the point $x_0$ as $k \to \infty$, so for $j \neq k$ sufficiently large, the supports are disjoint and $\|\phi_j - \phi_k\|_{L^{p^*}} = (\|\phi_j\|_{L^{p^*}}^{p^*} + \|\phi_k\|_{L^{p^*}}^{p^*})^{1/p^*} = 2^{1/p^*}$. No subsequence is Cauchy in $L^{p^*}(U)$, so the embedding is not compact. This is the same logical pattern as the Riesz Lemma argument: construct a bounded sequence with uniform separation, conclude non-compactness. The key difference is that the separation is achieved not by Riesz's Lemma directly, but by the concentration phenomenon — functions that concentrate at a point are automatically separated in $L^{p^*}$. [/example] ## Standard Arguments Using Riesz's Lemma The following techniques appear frequently in functional analysis and rely on Riesz's Lemma or its consequences as a key ingredient. Each illustrates a standard pattern of argument that the reader will encounter repeatedly. ### Constructing Separated Sequences The most direct application of Riesz's Lemma is the construction of sequences of unit vectors that are uniformly separated. The pattern is: 1. Start with a nested chain of finite-dimensional subspaces $Y_1 \subsetneq Y_2 \subsetneq \cdots \subsetneq X$. 2. At each step, apply Riesz's Lemma with a fixed $\theta \in (0,1)$ to find $x_k$ with $\|x_k\| = 1$ and $\operatorname{dist}(x_k, Y_k) \ge \theta$. 3. The resulting sequence satisfies $\|x_j - x_k\| \ge \theta$ for $j < k$, since $x_j \in Y_k$ for $j < k$. This construction is the prototype argument for *disproving* compactness, total boundedness, or the existence of convergent subsequences. Variations appear in: - **Proving non-compactness of bounded sets.** If a bounded set $S$ contains an infinite-dimensional affine subspace, apply Riesz's Lemma within the subspace to produce a separated sequence in $S$. - **Showing operators are not compact.** If $T \in \mathcal{L}(X, Y)$ and $T(x_k)$ remains separated (or has a separated subsequence), then $T$ is not compact. - **Bounding essential norms from below.** The essential norm $\|T\|_e := \inf\{\|T - K\| : K \text{ compact}\}$ satisfies $\|T\|_e \ge \limsup_k \|Tx_k\|$ for any $\theta$-separated sequence $(x_k)$ of unit vectors, since compact operators map such sequences to sequences that converge to $0$ (after passing to a subsequence). ### Closedness of Finite-Dimensional Subspaces An important ingredient in the Riesz Lemma construction — and a result that itself can be proved using Riesz's Lemma — is that every finite-dimensional subspace of a normed space is closed. [quotetheorem:1225] The standard proof uses the equivalence of all norms in finite dimensions. If $(y_k)_{k=1}^\infty \subset Y$ converges to some $x \in X$ in the norm of $X$, then $(y_k)$ is Cauchy in $X$, hence Cauchy in $Y$ (with the restricted norm). Since $\dim Y < \infty$, all norms on $Y$ are equivalent, so $(y_k)$ is Cauchy in $Y$ under any norm. In particular, it converges in $Y$ (every finite-dimensional normed space is complete), and the limit must be $x$ by uniqueness of limits in $X$. Therefore $x \in Y$, so $Y$ is closed. This fact is essential for the inductive construction in the non-compactness proof: at each step, the subspace $Y_k = \operatorname{span}\{x_1, \ldots, x_k\}$ must be closed for Riesz's Lemma to apply. Without closedness of finite-dimensional subspaces, the entire inductive machine would fail. ### The Contrapositive: Compact Ball Forces Finite Dimension In applications to operator theory, one frequently uses the contrapositive of the Riesz characterisation: if a subspace has a compact unit ball, it must be finite-dimensional. The two principal applications are: **Eigenspaces of compact operators.** If $T \in \mathcal{L}(X, X)$ is compact and $\lambda \neq 0$, then $\ker(T - \lambda I)$ is finite-dimensional. On $\ker(T - \lambda I)$, the identity satisfies $\operatorname{Id} = \lambda^{-1} T$, so the identity on this subspace is a scalar multiple of a compact operator, hence compact. The Riesz characterisation forces $\dim \ker(T - \lambda I) < \infty$. **Closed range of compact operators.** If $T \in \mathcal{L}(X, Y)$ is compact and $\operatorname{range}(T)$ is closed, then $\operatorname{range}(T)$ is finite-dimensional. Let $Z := \ker(T)$. The induced map $\tilde{T}: X/Z \to \operatorname{range}(T)$ is a bounded bijection, and by the open mapping theorem (assuming $X$ is a Banach space), $\tilde{T}^{-1}$ is bounded. Thus $\tilde{T}^{-1}(\overline{B}_{\operatorname{range}(T)}(0,1))$ is a bounded subset of $X/Z$, and its image under $\tilde{T}$ — the unit ball of $\operatorname{range}(T)$ — has compact closure (since $T$ is compact). The Riesz characterisation forces $\dim \operatorname{range}(T) < \infty$. These two observations are the starting point of **Riesz-Schauder theory**, which develops the spectral theory of compact operators on Banach spaces. The key results — that the spectrum of a compact operator consists of at most countably many eigenvalues accumulating only at $0$, and that each nonzero eigenvalue has finite multiplicity — all rely on the Riesz characterisation to pass from compactness conditions on operators to finite-dimensionality conditions on subspaces. ## References - F. Riesz, "Über lineare Funktionalgleichungen," *Acta Mathematica* **41** (1918), 71--98. - H. Brezis, *Functional Analysis, Sobolev Spaces and Partial Differential Equations* (2011). - J. Conway, *A Course in Functional Analysis* (1990). - E. Kreyszig, *Introductory Functional Analysis with Applications* (1978). - W. Rudin, *Functional Analysis* (1991). - P. Lax, *Functional Analysis* (2002).