1

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

Community
  • 1
Educ
  • 4,940
  • What do you mean with "sup" of a constant in the definition of $\varphi$? Also,$f,g$ are only defined on $[0,1]$, but it seems as if you are plugging in arbitrary real numbers. – PhoemueX Jan 11 '15 at 10:57
  • Can you specify over what set the supremum is over? If it is over $u \in \mathbb{R}$, then $\phi$ should not depend on $u$. – Empiricist Jan 11 '15 at 11:03
  • @PhoemueX it was just typo why you didn't fix it – Educ Jan 11 '15 at 11:36
  • @SRX i just forget to add supremum over $[0,1]$ – Educ Jan 11 '15 at 11:38
  • 1
    It was honestly not clear to me what you were asking. If you had written $x$ and $u$ instead of just $u$, I might have guessed. Now the question looks fine. – PhoemueX Jan 11 '15 at 13:08
  • @PhoemueX Thank You for Your Reviews – Educ Jan 11 '15 at 13:27

1 Answers1

3

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 $$ |\varphi(u)- \varphi(v)|\le M|u-v| \tag{4} $$

  • we have that $M=\sup_{[0,1]}|g|$ since g is bounded over $[0,1]$ and you add $|.|$ because we don't know if g is increasing or decreasing am i right – Educ Jan 11 '15 at 16:02
  • Yes, the supremum is finite because a continuous function on $[0,1]$ is bounded. I put absolute value because $g$ could take positive or negative values. It does not matter if it's increasing or decreasing. –  Jan 11 '15 at 16:14
  • they said only that are bounded over compact not continuous how did you know they are continuous – Educ Jan 11 '15 at 17:21
  • Okay, they are bounded and that's the only thing I used. Forget that I said "continuous". –  Jan 11 '15 at 17:57
  • but can we say that they are continuous – Educ Jan 11 '15 at 18:02
  • why would it be the same $x$ that gives sup for $u$ and $v$ – Educ Jan 11 '15 at 18:06
  • We can say they are continuous, but it would be a wrong thing to say. Also, nobody said the same $x$ would give $\sup$ for both $u$ and $v$. Also, this is my last comment here. –  Jan 11 '15 at 18:28