Let $f$ and $g$ two function bounded such that : $$f: [0,1] \to \mathbb{R}$$ $$g: [0,1] \to \mathbb{R}$$ Let : $$\varphi(u)=\sup_{x\in[0,1]}(f(x)+u\ g(x)),\quad u \in \mathbb{R}$$ Show that $\varphi$ is Lipschitz continuous.
We have to show that
$$\forall(x,y)\in R^2,~|\varphi(x)-\varphi(y)|\le k~|x-y|.$$
$f$ and $g$ are two bounded function then $\exists M\in \mathbb{R}$ such that $\forall x\in [0 ,1]$ we have $|f(x)| \leq M $ and $|g(x)| \leq M$. Let $u,v\in \mathbb{R}$. Let $x\in [0,1]$. we have : $$f(x) + u.g(x) = f(x) + v.g(x) + (u -v)g(x)\leq \varphi(v) + |u -v|.M$$
How can I goes to $\sup$ to say :
$$\varphi(u)\leq \varphi(v) + |u-v|.M$$.
Edit :
is that because of $f,g$ are bounded over compact $[0,1]$ then supremum of $f(x) + u.g(x)$ exists. but the problem is that $\sqrt{x}$ is bounded over compact $[0,1]$ and see this $\sqrt{x}$ isn't Lipschitz function
Notational confusion nonwithstanding, it is clear from the proof you began that the function you consider is $$\varphi(u)=\sup_{x\in[0,1]}(f(x)+u\ g(x)),\quad u \in \mathbb{R}$$ Following what you already did, let $M=\sup_{[0,1]}|g|$ (no need to worry about $f$). Fix $u$ and $v$ and note that for all $x\in [0,1]$ $$ f(x)+u\ g(x) \le (f(x)+v\ g(x))+M|u-v| \le \varphi(v)+M|u-v| \tag{1} $$ Formula (1) says that the number $\varphi(v)+M|u-v|$ is an upper bound for the set $\{f(x)+u\ g(x):x\in[0,1]\}$. The supremum, by definition, is the least upper bound. Hence, $$ \varphi(u)\le \varphi(v)+M|u-v| \tag{2} $$ Repeat the above with $u,v$ interchanged to get $$ \varphi(v)\le \varphi(u)+M|u-v| \tag{3} $$ Finally, combine (2) and (3) to obtain $$ |\varphi(u)- \varphi(v)|\le M|u-v| \tag{4} $$