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  "body": "\n## Proof\n\nWe prove both directions. The metrics are $\\|x - y\\|_V$ and $\\|x - y\\|_W$.\n\n1. Suppose $T$ is bounded, with $\\|Tx\\|_W \\le C \\|x\\|_V$ for all $x$. By linearity, $\\|Tx - Ty\\|_W = \\|T(x - y)\\|_W \\le C \\|x - y\\|_V$. So $T$ is [[Lipschitz|0TMZ21WKCN]] with constant $C$.\n2. Let $a \\in V$ and $\\varepsilon \u003e 0$, and let $\\delta = \\varepsilon / (C + 1)$. If $\\|x - a\\|_V \u003c \\delta$, then $\\|Tx - Ta\\|_W \\le C \\delta \u003c \\varepsilon$.\n3. By [[continuity via open sets|0YF82GBTND]], $T$ is continuous. So $T$ is a [[continuous linear map|0S3SMRH0Z5]].\n4. Conversely, suppose $T$ is continuous. The open ball $B = \\{w \\in W : \\|w\\|_W \u003c 1\\}$ is [[open|0B2305GNAZ]], so $T^{-1}(B)$ is open in $V$.\n5. Since $T0 = 0 \\in B$, we have $0 \\in T^{-1}(B)$. By [[the ball criterion|0A26X79RTB]], there is $\\eta \u003e 0$ with $\\|Tx\\|_W \u003c 1$ whenever $\\|x\\|_V \u003c \\eta$.\n6. Let $\\delta = \\eta / 2$, and let $x \\in V$ with $x \\neq 0$. Put $y = \\delta x / \\|x\\|_V$. Then $\\|y\\|_V = \\delta \u003c \\eta$, so $\\|Ty\\|_W \u003c 1$.\n7. By linearity, $Ty = (\\delta / \\|x\\|_V) \\, Tx$. So $\\|Tx\\|_W = (\\|x\\|_V / \\delta) \\|Ty\\|_W \u003c \\|x\\|_V / \\delta$.\n8. For $x = 0$, $Tx = 0$. So $\\|Tx\\|_W \\le C \\|x\\|_V$ for all $x \\in V$, with $C = 1 / \\delta$, and $T$ is bounded.\n\n## Idea\n\nA bound $\\|Tx\\| \\le C \\|x\\|$ makes $T$ Lipschitz, hence continuous. Conversely, continuity at $0$ gives a small ball that $T$ maps into the unit ball. Linearity lets us rescale any vector into that small ball, which yields the bound.\n",
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