[proofplan] We prove the result directly from the definitions of Lipschitz continuity and function composition. For arbitrary points $x_1,x_2\in X$, the $M$-Lipschitz estimate for $g$ controls the distance between $g(f(x_1))$ and $g(f(x_2))$ by the distance between $f(x_1)$ and $f(x_2)$ in $Y$. The $L$-Lipschitz estimate for $f$ then controls that intermediate distance by $d_X(x_1,x_2)$, and multiplying the two estimates gives the Lipschitz constant $ML$. [/proofplan]

[step:Declare the composition as a map from $X$ to $Z$] Since $f:X\to Y$ and $g:Y\to Z$, the composition \begin{align*} g\circ f:X\to Z \end{align*} is the map defined by \begin{align*} (g\circ f)(x)=g(f(x)) \end{align*} for every $x\in X$. [/step]

[step:Apply the two Lipschitz estimates in sequence]Let $x_1,x_2\in X$ be arbitrary. Since $f:X\to Y$, both $f(x_1)$ and $f(x_2)$ are elements of $Y$. Applying the $M$-Lipschitz property of $g:Y\to Z$ to the points $f(x_1),f(x_2)\in Y$ gives \begin{align*} d_Z(g(f(x_1)),g(f(x_2)))\le M\,d_Y(f(x_1),f(x_2)). \end{align*} Applying the $L$-Lipschitz property of $f:X\to Y$ to the points $x_1,x_2\in X$ gives \begin{align*} d_Y(f(x_1),f(x_2))\le L\,d_X(x_1,x_2). \end{align*} Because $M\ge 0$, multiplying the [second inequality](/theorems/2136) by $M$ preserves the inequality: \begin{align*} M\,d_Y(f(x_1),f(x_2))\le ML\,d_X(x_1,x_2). \end{align*} Combining the preceding two inequalities yields \begin{align*} d_Z(g(f(x_1)),g(f(x_2)))\le ML\,d_X(x_1,x_2). \end{align*}[/step]

[guided]We start with arbitrary points $x_1,x_2\in X$ because the definition of being Lipschitz requires an estimate for every pair of points in the domain. The composition $g\circ f$ sends $x\in X$ first to $f(x)\in Y$ and then to $g(f(x))\in Z$, so the distance we must estimate is \begin{align*} d_Z((g\circ f)(x_1),(g\circ f)(x_2)). \end{align*} By the definition of composition, this is \begin{align*} d_Z(g(f(x_1)),g(f(x_2))). \end{align*} The map $g:Y\to Z$ is $M$-Lipschitz, meaning that for every $y_1,y_2\in Y$, \begin{align*} d_Z(g(y_1),g(y_2))\le M\,d_Y(y_1,y_2). \end{align*} We may apply this with $y_1=f(x_1)$ and $y_2=f(x_2)$ because $f$ takes values in $Y$. Therefore, \begin{align*} d_Z(g(f(x_1)),g(f(x_2)))\le M\,d_Y(f(x_1),f(x_2)). \end{align*} Now we estimate the remaining distance in $Y$. The map $f:X\to Y$ is $L$-Lipschitz, meaning that for every $u_1,u_2\in X$, \begin{align*} d_Y(f(u_1),f(u_2))\le L\,d_X(u_1,u_2). \end{align*} Applying this with $u_1=x_1$ and $u_2=x_2$ gives \begin{align*} d_Y(f(x_1),f(x_2))\le L\,d_X(x_1,x_2). \end{align*} Since $M\ge 0$, multiplying this inequality by $M$ preserves the direction of the inequality: \begin{align*} M\,d_Y(f(x_1),f(x_2))\le ML\,d_X(x_1,x_2). \end{align*} Substituting this bound into the earlier estimate gives \begin{align*} d_Z(g(f(x_1)),g(f(x_2)))\le ML\,d_X(x_1,x_2). \end{align*} This is precisely the desired Lipschitz estimate for $g\circ f$ at the pair $x_1,x_2$.[/guided]

[step:Conclude the Lipschitz bound for every pair of points] Using $(g\circ f)(x_i)=g(f(x_i))$ for $i\in\{1,2\}$, the estimate from the previous step becomes \begin{align*} d_Z((g\circ f)(x_1),(g\circ f)(x_2))\le ML\,d_X(x_1,x_2). \end{align*} Since $x_1,x_2\in X$ were arbitrary, $g\circ f:X\to Z$ is $ML$-Lipschitz. [/step]