{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe apply the mean value inequality to $f$ minus its linear part at $a$.\n\n1. Let $U$ and $f$ be as in the statement, and let $a \\in U$. Write $\\norm{B}$ for the [[operator norm|0ZTEH055YF]] of a matrix $B$.\n2. Since $f$ is of [[class C^1|0TCBZMEYGN]], it is [[differentiable|0KXSK8TEEB]] at each point $z \\in U$, and $Df(z)$ is the [[Jacobian matrix|0YF0VXKXGB]] of partial derivatives at $z$.\n3. For a matrix $B = (b_{ij})$, the [[Cauchy–Schwarz inequality|0RFWW56Y22]] on each row gives $\\norm{B} \\le (\\sum_{i, j} b_{ij}^2)^{1/2}$.\n4. Let $\\varepsilon \u003e 0$. The partial derivatives of $f$ are continuous at $a$. By step 3, there is $r \u003e 0$ such that the ball $B_r = \\set{z : \\norm{z - a} \u003c r}$ lies in $U$ and $\\norm{Df(z) - Df(a)} \\le \\varepsilon$ for $z \\in B_r$.\n5. Let $g(z) = f(z) - Df(a) z$ for $z \\in U$. A linear map is its own derivative, so $g$ is differentiable on $U$ with $Dg(z) = Df(z) - Df(a)$.\n6. The ball $B_r$ is [[convex|0K06X9QXDX]], so it contains the segment between any two of its points. By the [[mean value inequality|02CYB184RC]] and step 4, $\\norm{g(x) - g(y)} \\le \\varepsilon \\norm{x - y}$ for $x, y \\in B_r$.\n7. Since $g(x) - g(y) = f(x) - f(y) - Df(a)(x - y)$, this says $\\norm{f(x) - f(y) - Df(a)(x - y)} \\le \\varepsilon \\norm{x - y}$ for $x, y \\in B_r$.\n8. If $\\norm{(x, y) - (a, a)} \u003c r$, then $\\norm{x - a} \u003c r$ and $\\norm{y - a} \u003c r$, so $x, y \\in B_r$. For such pairs with $x \\ne y$, step 7 gives $\\frac{\\norm{f(x) - f(y) - Df(a)(x - y)}}{\\norm{x - y}} \\le \\varepsilon$.\n9. Since $\\varepsilon \u003e 0$ was arbitrary, this quotient tends to $0$ as $(x, y) \\to (a, a)$ with $x \\ne y$. So $f$ is [[strictly differentiable|0146MBQNNE]] at $a$, with linear map $Df(a)$.\n\n## Idea\n\nSubtracting the linear map $Df(a)$ leaves a function whose derivative is small near $a$, because $Df$ is continuous. The mean value inequality turns a small derivative into a small Lipschitz constant, and this is exactly the estimate that strict differentiability asks for.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "26B05"
    ],
    "name": "Proof that C^1 functions are strictly differentiable",
    "proves": "0STPPDXGME",
    "sources": [
      {
        "locator": "Basic definition",
        "url": "https://en.wikipedia.org/wiki/Strict_differentiability"
      }
    ],
    "tag": "HDMDRA7ZBW",
    "uses": [
      "0146MBQNNE",
      "0TCBZMEYGN",
      "0KXSK8TEEB",
      "0YF0VXKXGB",
      "0ZTEH055YF",
      "0RFWW56Y22",
      "02CYB184RC",
      "0K06X9QXDX"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/HDMDRA7ZBW/",
  "links": [
    {
      "field": "proves",
      "target": "0STPPDXGME"
    },
    {
      "field": "uses",
      "target": "0146MBQNNE"
    },
    {
      "field": "uses",
      "target": "0TCBZMEYGN"
    },
    {
      "field": "uses",
      "target": "0KXSK8TEEB"
    },
    {
      "field": "uses",
      "target": "0YF0VXKXGB"
    },
    {
      "field": "uses",
      "target": "0ZTEH055YF"
    },
    {
      "field": "uses",
      "target": "0RFWW56Y22"
    },
    {
      "field": "uses",
      "target": "02CYB184RC"
    },
    {
      "field": "uses",
      "target": "0K06X9QXDX"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/HDMDRA7ZBW.md",
  "msc": [
    "26B05"
  ],
  "name": "Proof that C^1 functions are strictly differentiable",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/HDM/HDMDRA7ZBW.md",
  "status": null,
  "tag": "HDMDRA7ZBW",
  "uri": "m2://HDMDRA7ZBW",
  "url": "https://math2.tamnd.com/wiki/proof-that-c-1-functions-are-strictly-differentiable/",
  "writing": {
    "tag": "HDMDRA7ZBW",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}