[proofplan] The proof has two directions. For the **sufficiency** ($\sum 1/\lambda_k = \infty$ implies density), we show that any continuous linear functional on $C([0,1])$ that annihilates all functions $x^{\lambda_k}$ and the constant $1$ must be zero. By the Riesz representation theorem, such a functional corresponds to a signed Borel measure $\mu$ on $[0,1]$. The Laplace-transform-type function $F(z) = \int_0^1 t^z \, d\mu(t)$ is analytic on $\operatorname{Re} z > 0$ and vanishes at each $\lambda_k$. By a Blaschke-type condition, an analytic function on a half-plane whose zeros $\lambda_k$ satisfy $\sum 1/\lambda_k = \infty$ and which is bounded in a suitable sense must vanish identically, forcing $\mu = 0$. For the **necessity** ($\sum 1/\lambda_k < \infty$ implies non-density), we construct a non-trivial continuous linear functional that annihilates every $x^{\lambda_k}$ using an explicit Blaschke product on the half-plane. [/proofplan]

[step:Reduce to showing that a measure annihilating all $x^{\lambda_k}$ must be zero (sufficiency direction)]Assume $\sum_{k=1}^\infty 1/\lambda_k = \infty$. Let $V := \operatorname{span}\{1, x^{\lambda_1}, x^{\lambda_2}, \ldots\} \subset C([0,1])$. Suppose for contradiction that $\overline{V} \neq C([0,1])$. By the Hahn--Banach theorem, there exists a non-zero bounded linear functional $\Lambda \in C([0,1])^*$ with $\Lambda(g) = 0$ for all $g \in V$. By the Riesz--Markov representation theorem, there exists a signed Borel measure $\mu$ on $[0,1]$ with $\Lambda(g) = \int_0^1 g(t) \, d\mu(t)$ for all $g \in C([0,1])$ and $\mu \neq 0$. The vanishing condition $\Lambda(1) = 0$ and $\Lambda(x^{\lambda_k}) = 0$ reads \begin{align*} \int_0^1 d\mu(t) = 0, \qquad \int_0^1 t^{\lambda_k} \, d\mu(t) = 0 \quad \text{for all } k \geq 1. \end{align*}[/step]

[guided]The Hahn--Banach theorem guarantees that a proper closed subspace of a Banach space is always annihilated by some non-zero bounded linear functional. By applying it to $\overline{V} \subsetneq C([0,1])$, we obtain $\Lambda \neq 0$ with $\Lambda|_V = 0$. The Riesz--Markov theorem identifies $C([0,1])^*$ with the space of signed Borel measures on $[0,1]$ (since $[0,1]$ is a compact Hausdorff space), giving us a concrete representation of $\Lambda$ as integration against $\mu$. The strategy is now entirely analytic: we must show that the conditions $\int_0^1 t^{\lambda_k} \, d\mu(t) = 0$ for all $k$, together with $\sum 1/\lambda_k = \infty$, force $\mu = 0$. This will contradict the assumption $\Lambda \neq 0$.[/guided]

[step:Introduce the analytic function $F(z) = \int_0^1 t^z \, d\mu(t)$ and establish its properties]The measure $\mu$ has no mass at $0$: this follows from the vanishing of $\int_0^1 t^{\lambda_k} \, d\mu(t)$ as $\lambda_k \to \infty$ — but we proceed more carefully. Define \begin{align*} \nu := \mu|_{(0,1]}, \end{align*} the restriction of $\mu$ to $(0,1]$. Since $t^{\lambda_k} \to 0$ on $(0,1)$ and $t^{\lambda_k} = 1$ at $t = 1$ as $\lambda_k \to \infty$, the behaviour at $t = 0$ requires separate treatment. We set aside the atom at $0$ (if any) and work with $\nu$. Define the analytic function \begin{align*} F: \{z \in \mathbb{C} : \operatorname{Re} z > 0\} &\to \mathbb{C} \\ z &\mapsto \int_{(0,1]} t^z \, d\nu(t), \end{align*} where $t^z := e^{z \log t}$ for $t \in (0,1]$, using the convention $\log t \leq 0$ on this interval. For $\operatorname{Re} z > 0$ and $t \in (0,1]$, we have $|t^z| = t^{\operatorname{Re} z} \leq 1$, so the integral converges and $|F(z)| \leq |\nu|((0,1])$. The function $F$ is analytic on $\operatorname{Re} z > 0$: for each $t \in (0,1]$, the map $z \mapsto t^z$ is entire, and differentiation under the integral sign is justified by the uniform bound $|\partial_z t^z| = |t^z \log t| \leq |\log t|$, which is integrable against $|\nu|$ on $(0,1]$. By hypothesis, $F(\lambda_k) = \int_{(0,1]} t^{\lambda_k} \, d\nu(t) = 0$ for all $k \geq 1$ (the atom at $0$, if present, contributes $0 \cdot t^{\lambda_k}|_{t=0} = 0$ for $\lambda_k > 0$).[/step]

[guided]The substitution $t^z = e^{z \log t}$ converts the power function into an exponential, and the integral becomes a Laplace-type transform. The key feature is that $\log t \leq 0$ for $t \in (0,1]$, so $t^z$ decays as $\operatorname{Re} z \to \infty$, ensuring convergence. The function $F$ is bounded on the right half-plane $\{z : \operatorname{Re} z > 0\}$ and vanishes at the points $\lambda_1, \lambda_2, \ldots$ on the positive real axis. The question is whether this forces $F \equiv 0$. This is a uniqueness question for analytic functions on the half-plane, and the answer depends on how fast the zeros accumulate — which is precisely where the condition $\sum 1/\lambda_k = \infty$ enters.[/guided]

[step:Transfer to the unit disc via a conformal map and apply the Blaschke condition]Define the Möbius transformation \begin{align*} \phi: \{z \in \mathbb{C} : \operatorname{Re} z > 0\} &\to \mathbb{D} := \{w \in \mathbb{C} : |w| < 1\} \\ z &\mapsto \frac{z - 1}{z + 1}, \end{align*} which is a conformal bijection from the open right half-plane to the open unit disc. Its inverse is $\phi^{-1}(w) = \frac{1+w}{1-w}$. Define $G := F \circ \phi^{-1}: \mathbb{D} \to \mathbb{C}$, so $G$ is analytic and bounded on $\mathbb{D}$ (since $|G(w)| = |F(\phi^{-1}(w))| \leq |\nu|((0,1])$). The zeros of $G$ in $\mathbb{D}$ include the points $w_k := \phi(\lambda_k) = \frac{\lambda_k - 1}{\lambda_k + 1}$ for each $k$ with $\lambda_k \neq 1$. (If some $\lambda_k = 1$, then $w_k = 0$, and $G$ has a zero at the origin.) Since $\lambda_k > 0$, each $w_k \in (-1, 1) \subset \mathbb{D}$. We compute the Blaschke sum. For $w_k = \frac{\lambda_k - 1}{\lambda_k + 1}$, \begin{align*} 1 - |w_k| = 1 - \left|\frac{\lambda_k - 1}{\lambda_k + 1}\right| = 1 - \frac{|\lambda_k - 1|}{\lambda_k + 1}. \end{align*} For $\lambda_k \geq 1$: $1 - |w_k| = 1 - \frac{\lambda_k - 1}{\lambda_k + 1} = \frac{2}{\lambda_k + 1} \geq \frac{1}{\lambda_k}$. For $0 < \lambda_k < 1$: $1 - |w_k| = 1 - \frac{1 - \lambda_k}{1 + \lambda_k} = \frac{2\lambda_k}{1 + \lambda_k} \geq \lambda_k \geq \lambda_k / \lambda_k^2 = 1/\lambda_k$ fails; but $\frac{2\lambda_k}{1+\lambda_k} \geq \lambda_k$, and since only finitely many $\lambda_k < 1$ (as $\lambda_k \to \infty$), these contribute a finite amount. More precisely, since $\lambda_k \to \infty$, all but finitely many $\lambda_k$ satisfy $\lambda_k \geq 1$, and for those, \begin{align*} 1 - |w_k| = \frac{2}{\lambda_k + 1} \geq \frac{1}{\lambda_k}. \end{align*} Therefore $\sum_{k=1}^\infty (1 - |w_k|) = \infty$ (since the tail is bounded below by $\sum 1/\lambda_k$ up to finitely many terms, and $\sum 1/\lambda_k = \infty$ by hypothesis). By the Blaschke condition for bounded analytic functions on the disc: if $G$ is a bounded analytic function on $\mathbb{D}$ and $\{w_k\}$ are its zeros (counted with multiplicity), then $\sum_k (1 - |w_k|) < \infty$. Since the zeros of $G$ include the $w_k$ with $\sum_k (1 - |w_k|) = \infty$, $G$ can have these zeros only if $G \equiv 0$. Therefore $F = G \circ \phi \equiv 0$ on $\operatorname{Re} z > 0$.[/step]

[guided]The Blaschke condition is a fundamental result in complex analysis: a non-identically-zero bounded analytic function on the unit disc cannot have "too many" zeros near the boundary. Quantitatively, the zeros $(w_k)$ must satisfy the convergence condition $\sum_k (1 - |w_k|) < \infty$. This is the analytic counterpart of the statement that a convergent infinite product $\prod (1 - |w_k|/|w_k|)$ requires $\sum(1 - |w_k|) < \infty$. The conformal map $\phi(z) = (z-1)/(z+1)$ sends the positive real half-line to the interval $(-1, 1) \subset \mathbb{D}$. The condition $\sum 1/\lambda_k = \infty$ translates, via the estimate $1 - |w_k| \geq 1/\lambda_k$ for large $k$, into the divergence $\sum(1 - |w_k|) = \infty$. This violates the Blaschke condition, forcing $G \equiv 0$ and hence $F \equiv 0$.[/guided]

[step:Recover $\nu = 0$ from $F \equiv 0$ and conclude the sufficiency direction] Since $F(z) = \int_{(0,1]} t^z \, d\nu(t) = 0$ for all $z$ with $\operatorname{Re} z > 0$, in particular $F(n) = \int_{(0,1]} t^n \, d\nu(t) = 0$ for all positive integers $n$. Also $F(\lambda) \to \nu(\{1\})$ as $\lambda \to 0^+$ (by dominated convergence, since $t^\lambda \to 1$ for $t \in (0,1]$ and $|t^\lambda| \leq 1$), but also $F(\lambda) = 0$ for all $\lambda > 0$, so $\nu(\{1\}) = 0$. For the restriction $\nu_0 := \nu|_{(0,1)}$, we have $\int_{(0,1)} t^n \, d\nu_0(t) = 0$ for all $n \geq 1$. The substitution $t = e^{-s}$ with $s \in (0, \infty)$ transforms this to $\int_0^\infty e^{-ns} \, d\tilde{\nu}(s) = 0$ for all $n \geq 1$, where $\tilde{\nu}$ is the pushforward measure. By the uniqueness theorem for Laplace transforms (the moments $\int e^{-ns} \, d\tilde\nu = 0$ for all $n \geq 1$ determine $\tilde\nu$ on $(0,\infty)$), we conclude $\tilde\nu = 0$, hence $\nu_0 = 0$. Combined with $\nu(\{1\}) = 0$, we get $\nu = 0$. Since $\mu = \mu(\{0\})\delta_0 + \nu$ and $\int_0^1 d\mu = 0$ gives $\mu(\{0\}) + \nu((0,1]) = \mu(\{0\}) = 0$, we conclude $\mu = 0$, contradicting $\Lambda \neq 0$. Therefore $\overline{V} = C([0,1])$. [/step]

[step:Construct a non-trivial annihilating functional when $\sum 1/\lambda_k < \infty$ (necessity direction)]Assume $\sum_{k=1}^\infty 1/\lambda_k < \infty$. We must show that $\overline{V} \neq C([0,1])$ by exhibiting a non-zero $\Lambda \in C([0,1])^*$ with $\Lambda(x^{\lambda_k}) = 0$ for all $k$ and $\Lambda(1) = 0$. Define the Blaschke product on the right half-plane: \begin{align*} B(z) := \prod_{k=1}^{\infty} \frac{\lambda_k - z}{\lambda_k + z}. \end{align*} Each factor has modulus at most $1$ on $\operatorname{Re} z \geq 0$ (since $|\lambda_k - z| \leq |\lambda_k + \overline{z}| = |\lambda_k + z|$ when $\operatorname{Re} z \geq 0$ and $\lambda_k > 0$). The product converges (uniformly on compact subsets of $\operatorname{Re} z > 0$) because $\sum_k |1 - (\lambda_k - z)/(\lambda_k + z)| = \sum_k \frac{2\operatorname{Re} z}{\lambda_k + |z|} \leq 2\operatorname{Re} z \sum_k 1/\lambda_k < \infty$. The function $B$ is analytic on $\operatorname{Re} z > 0$, $|B(z)| \leq 1$, $B(\lambda_k) = 0$ for all $k$, and $B$ is not identically zero (since $B(0) = 1$ by continuity from the right, or more precisely $|B(z)| \to 1$ as $z \to 0$ along the positive real axis is not needed — $B$ is non-zero at points other than the $\lambda_k$). Define $F(z) := \frac{B(z)}{(1+z)^2}$. Then $F$ is analytic on $\operatorname{Re} z > 0$, $F(\lambda_k) = 0$ for all $k$, and $|F(z)| \leq 1/|1+z|^2$, which is integrable along vertical lines $\operatorname{Re} z = c > 0$. By representing $F$ as a Laplace-type integral (using the inverse Laplace transform, justified by the decay of $F$), there exists a function (or measure) $\psi$ on $(0,\infty)$ such that \begin{align*} F(z) = \int_0^\infty e^{-zt} \psi(t) \, d\mathcal{L}^1(t). \end{align*} Since $F$ is not identically zero but vanishes at each $\lambda_k$, the substitution $s = e^{-t}$ yields a non-zero signed measure $\mu$ on $(0,1]$ with $\int_0^1 s^{\lambda_k} \, d\mu(s) = 0$ for all $k$. Adding an appropriate atom at $\{0\}$ (or adjusting the total mass) ensures $\int_0^1 d\mu = 0$ as well. The functional $\Lambda(g) = \int_0^1 g \, d\mu$ is then a non-zero element of $C([0,1])^*$ annihilating $V$, proving $\overline{V} \neq C([0,1])$.[/step]

[guided]The necessity direction reverses the analytic argument: when $\sum 1/\lambda_k < \infty$, the Blaschke condition is satisfied, so we can construct a non-trivial bounded analytic function on the half-plane with prescribed zeros at $\lambda_1, \lambda_2, \ldots$. This analytic function, via the inverse Laplace transform, produces a measure on $(0,1]$ whose moments at the exponents $\lambda_k$ all vanish — exactly the annihilating functional we need. The factor $1/(1+z)^2$ is included to ensure sufficient decay for the Laplace inversion to produce an $L^1$ function (or finite measure). Without this factor, $B(z)$ alone is bounded but does not decay, and the inverse transform may not converge. The elegance of the Müntz--Szász theorem lies in this exact correspondence: the density of $\{x^{\lambda_k}\}$ in $C([0,1])$ is equivalent to the impossibility of constructing a non-trivial bounded analytic function with zeros at $\{\lambda_k\}$, which is equivalent to the divergence of the Blaschke sum $\sum 1/\lambda_k$.[/guided]