{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe move the problem to $[0, 1]$ and show that the Bernstein polynomials of the moved function approximate it uniformly.\n\n*Part 1: reduction to $[0, 1]$.*\n\n1. Let $a \u003c b$, let $f \\colon [a, b] \\to \\R$ be continuous, and let $\\varepsilon \u003e 0$. Put $L = b - a \u003e 0$.\n2. For $t \\in [0, 1]$, let $g(t) = f(a + tL)$. The map $t \\mapsto a + tL$ is continuous and sends $[0, 1]$ into $[a, b]$.\n3. So $g \\colon [0, 1] \\to \\R$ is continuous, since a [[composition of continuous functions is continuous|0NRQPQX3VA]].\n4. By the [[boundedness theorem|0ATVXC3NCC]], there is $M \u003e 0$ with $\\abs{g(t)} \\le M$ for all $t \\in [0, 1]$.\n5. Continuous functions on a closed bounded interval are [[uniformly continuous|0W0TZDRF9C]]. By the definition of [[uniform continuity|0A3CK6DVVB]], there is $\\delta \u003e 0$ with this property.\n$$s, t \\in [0, 1], \\ \\abs{s - t} \u003c \\delta \\implies \\abs{g(s) - g(t)} \u003c \\frac{\\varepsilon}{2}.$$\n\n*Part 2: three identities.*\n\n6. Fix an integer $n \\ge 1$ and $x \\in [0, 1]$. For $0 \\le k \\le n$, let $b_k = \\binom{n}{k} x^k (1 - x)^{n-k}$, the [[Bernstein basis polynomials|0C4VWBFY73]] at $x$. Each $b_k \\ge 0$.\n7. By the [[binomial theorem|0EHWPWGHB2]], $\\sum_{k=0}^{n} b_k = (x + (1 - x))^n = 1$.\n8. For $1 \\le k \\le n$, the [[factorial formula|033KTKKRV6]] gives $k \\binom{n}{k} = \\frac{n!}{(k-1)! \\, (n-k)!} = n \\binom{n-1}{k-1}$.\n9. The term with $k = 0$ vanishes. So by step 8, with $j = k - 1$ and the binomial theorem,\n$$\\sum_{k=0}^{n} k b_k = n x \\sum_{j=0}^{n-1} \\binom{n-1}{j} x^j (1 - x)^{n-1-j} = n x.$$\n10. Similarly, for $2 \\le k \\le n$, the factorial formula gives $k (k - 1) \\binom{n}{k} = n (n - 1) \\binom{n-2}{k-2}$.\n11. The terms with $k = 0$ and $k = 1$ vanish. If $n \\ge 2$, step 10 and the binomial theorem give $\\sum_{k=0}^{n} k (k - 1) b_k = n (n - 1) x^2$.\n12. If $n = 1$, both sides of this identity are $0$. So it holds for all $n \\ge 1$.\n13. Write $(k - nx)^2 = k (k - 1) + k - 2nx \\, k + n^2 x^2$. By steps 7, 9 and 11,\n$$\\sum_{k=0}^{n} (k - nx)^2 b_k = n(n-1) x^2 + nx - 2 n^2 x^2 + n^2 x^2 = n x (1 - x).$$\n14. Since $\\frac{1}{4} - x + x^2 = (x - \\frac{1}{2})^2 \\ge 0$, we have $x (1 - x) \\le \\frac{1}{4}$. Dividing step 13 by $n^2$ gives\n$$\\sum_{k=0}^{n} \\Big( \\frac{k}{n} - x \\Big)^2 b_k \\le \\frac{1}{4n}.$$\n\n*Part 3: the estimate.*\n\n15. The [[Bernstein polynomial|0C4VWBFY73]] of $g$ is $B_n(g)(x) = \\sum_{k=0}^{n} g(\\frac{k}{n}) b_k$. By step 7, $g(x) = \\sum_{k=0}^{n} g(x) b_k$.\n16. Subtracting, and using the [[triangle inequality|03HF5PMY5T]] with $b_k \\ge 0$,\n$$\\abs{g(x) - B_n(g)(x)} \\le \\sum_{k=0}^{n} \\Big| g(x) - g\\Big(\\frac{k}{n}\\Big) \\Big| \\, b_k.$$\n17. Let $K_1$ be the set of $k$ with $\\abs{\\frac{k}{n} - x} \u003c \\delta$. Let $K_2$ be the set of the remaining $k \\in \\set{0, \\dots, n}$.\n18. For $k \\in K_1$, step 5 gives $\\abs{g(x) - g(\\frac{k}{n})} \u003c \\frac{\\varepsilon}{2}$. So by step 7, the sum over $K_1$ is at most $\\frac{\\varepsilon}{2} \\sum_{k} b_k = \\frac{\\varepsilon}{2}$.\n19. For $k \\in K_2$, step 4 gives $\\abs{g(x) - g(\\frac{k}{n})} \\le 2M$. Also $1 \\le (\\frac{k}{n} - x)^2 / \\delta^2$.\n20. So by step 14, the sum over $K_2$ is at most\n$$\\frac{2M}{\\delta^2} \\sum_{k \\in K_2} \\Big( \\frac{k}{n} - x \\Big)^2 b_k \\le \\frac{2M}{\\delta^2} \\cdot \\frac{1}{4n} = \\frac{M}{2 n \\delta^2}.$$\n21. By the [[Archimedean property|0A74SX55QP]], choose an integer $n \\ge 1$ with $n \u003e \\frac{M}{\\varepsilon \\delta^2}$. Then $\\frac{M}{2 n \\delta^2} \u003c \\frac{\\varepsilon}{2}$.\n22. For this $n$, steps 16 to 21 give $\\abs{g(x) - B_n(g)(x)} \u003c \\frac{\\varepsilon}{2} + \\frac{\\varepsilon}{2} = \\varepsilon$. This holds for every $x \\in [0, 1]$, since $n$ does not depend on $x$.\n\n*Part 4: back to $[a, b]$.*\n\n23. Let $q = B_n(g)$, a polynomial in one variable. Define $p(x) = q\\big( \\frac{x - a}{L} \\big)$ for real $x$.\n24. Each power $\\big( \\frac{x - a}{L} \\big)^j$ expands into a polynomial in $x$, by the binomial theorem. So $p$ is a polynomial.\n25. Let $x \\in [a, b]$, and put $t = \\frac{x - a}{L}$. Then $t \\in [0, 1]$, $a + tL = x$, and so $g(t) = f(x)$.\n26. By step 22, $\\abs{f(x) - p(x)} = \\abs{g(t) - q(t)} \u003c \\varepsilon$. This proves the theorem.\n\n## Idea\n\nThe Bernstein polynomial $B_n(g)(x)$ is a weighted average of the values $g(k/n)$. The weights $b_k(x)$ concentrate near $k/n = x$, because their variance about $x$ is at most $1/(4n)$. Uniform continuity controls the nearby terms, and the variance bound makes the distant terms small.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "41A10"
    ],
    "name": "Proof of the Weierstrass approximation theorem",
    "proves": "0JQE130DHW",
    "sources": [
      {
        "locator": "Approximating continuous functions, Elementary proof",
        "url": "https://en.wikipedia.org/wiki/Bernstein_polynomial"
      }
    ],
    "tag": "BAVZZ1KJSY",
    "uses": [
      "0NRQPQX3VA",
      "0ATVXC3NCC",
      "0W0TZDRF9C",
      "0A3CK6DVVB",
      "0C4VWBFY73",
      "0EHWPWGHB2",
      "033KTKKRV6",
      "03HF5PMY5T",
      "0A74SX55QP"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/BAVZZ1KJSY/",
  "links": [
    {
      "field": "proves",
      "target": "0JQE130DHW"
    },
    {
      "field": "uses",
      "target": "0NRQPQX3VA"
    },
    {
      "field": "uses",
      "target": "0ATVXC3NCC"
    },
    {
      "field": "uses",
      "target": "0W0TZDRF9C"
    },
    {
      "field": "uses",
      "target": "0A3CK6DVVB"
    },
    {
      "field": "uses",
      "target": "0C4VWBFY73"
    },
    {
      "field": "uses",
      "target": "0EHWPWGHB2"
    },
    {
      "field": "uses",
      "target": "033KTKKRV6"
    },
    {
      "field": "uses",
      "target": "03HF5PMY5T"
    },
    {
      "field": "uses",
      "target": "0A74SX55QP"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/BAVZZ1KJSY.md",
  "msc": [
    "41A10"
  ],
  "name": "Proof of the Weierstrass approximation theorem",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/BAV/BAVZZ1KJSY.md",
  "status": null,
  "tag": "BAVZZ1KJSY",
  "uri": "m2://BAVZZ1KJSY",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-weierstrass-approximation-theorem/",
  "writing": {
    "tag": "BAVZZ1KJSY",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}