[proofplan] The proof is pointwise. First we verify that the composed functions are bounded, so the [uniform norm](/page/Uniform%20Norm) on the left is defined. Then, for each $x \in E$, the Lipschitz condition is applied to the two points $f(x),g(x) \in I$. Taking the supremum over $E$ converts this pointwise estimate into the desired uniform norm estimate. [/proofplan] [step:Verify that the composed functions are bounded] If $E=\varnothing$, then every function on $E$ is bounded and there is nothing to verify. Assume $E \ne \varnothing$, and choose $x_0 \in E$. Since $f: E \to I$ is bounded, there exists $M_f \ge 0$ such that $|f(x)| \le M_f$ for every $x \in E$. For every $x \in E$, the points $f(x)$ and $f(x_0)$ lie in $I$, so the Lipschitz condition gives \begin{align*} |\Phi(f(x))-\Phi(f(x_0))| \le L|f(x)-f(x_0)|. \end{align*} Using the triangle inequality and the bound on $f$, we obtain \begin{align*} |\Phi(f(x))| \le |\Phi(f(x_0))| + L|f(x)-f(x_0)|. \end{align*} Also, \begin{align*} |f(x)-f(x_0)| \le |f(x)|+|f(x_0)| \le 2M_f. \end{align*} Therefore \begin{align*} |\Phi(f(x))| \le |\Phi(f(x_0))|+2LM_f \end{align*} for all $x \in E$, so $\Phi \circ f: E \to \mathbb{R}$ is bounded. The same argument applied to $g: E \to I$ shows that $\Phi \circ g: E \to \mathbb{R}$ is bounded. Hence $\Phi \circ f-\Phi \circ g: E \to \mathbb{R}$ is bounded. [/step] [step:Apply the Lipschitz estimate pointwise and take the supremum] For each $x \in E$, the values $f(x)$ and $g(x)$ belong to $I$. Applying the assumed Lipschitz estimate to $s=f(x)$ and $t=g(x)$ gives \begin{align*} |\Phi(f(x))-\Phi(g(x))| \le L|f(x)-g(x)|. \end{align*} By the definition of the uniform norm, \begin{align*} |f(x)-g(x)| \le \|f-g\|_\infty \end{align*} for every $x \in E$. Combining the two inequalities gives \begin{align*} |\Phi(f(x))-\Phi(g(x))| \le L\|f-g\|_\infty \end{align*} for every $x \in E$. Taking the supremum over $x \in E$ yields \begin{align*} \|\Phi \circ f-\Phi \circ g\|_\infty \le L\|f-g\|_\infty. \end{align*} [guided] We want to prove an estimate in the uniform norm, so the natural first move is to prove the corresponding estimate at an arbitrary point of the domain. Fix $x \in E$. Since $f: E \to I$ and $g: E \to I$, both values $f(x)$ and $g(x)$ lie in the interval $I$. This is exactly the hypothesis needed to use the Lipschitz condition for $\Phi: I \to \mathbb{R}$. Applying that Lipschitz condition with $s=f(x)$ and $t=g(x)$ gives \begin{align*} |\Phi(f(x))-\Phi(g(x))| \le L|f(x)-g(x)|. \end{align*} The remaining step is to replace the pointwise quantity $|f(x)-g(x)|$ by the largest possible such quantity over all points of $E$. By definition, the uniform norm of the bounded function $f-g: E \to \mathbb{R}$ is \begin{align*} \|f-g\|_\infty = \sup_{y \in E}|f(y)-g(y)|. \end{align*} Therefore, for the particular point $x \in E$, \begin{align*} |f(x)-g(x)| \le \|f-g\|_\infty. \end{align*} Substituting this into the pointwise Lipschitz estimate gives \begin{align*} |\Phi(f(x))-\Phi(g(x))| \le L\|f-g\|_\infty. \end{align*} This bound holds for every $x \in E$. Taking the supremum over all $x \in E$ gives \begin{align*} \sup_{x \in E}|\Phi(f(x))-\Phi(g(x))| \le L\|f-g\|_\infty. \end{align*} The left-hand side is precisely $\|\Phi \circ f-\Phi \circ g\|_\infty$, so \begin{align*} \|\Phi \circ f-\Phi \circ g\|_\infty \le L\|f-g\|_\infty. \end{align*} [/guided] [/step]