Ytukyg

Today during an exam I got the following exercise :

Let $h, g in C^0([0,1], | cdot |_1)$ such that the set : $Z(g) = {x in [0,1] mid g(x) = 0}$ is a finite union of intervals. Then let's defined :
$$ G : t mapsto | g + th |_1$$

If the function $G$ has a derivative at $0$ prove that :
$$G'(0) = int_{[0,1] backslash Z(g)} h(x) cdot sign(g(x)) mathrm{d}x$$

First of all I don't understand how $G$ is defined because do we consider that :
$$G(t) = int_{[0,1]} mid g(x) + th(x) mid mathrm{d}x$$

Or we consider that :

$$G(t) = int_{[0,1]} mid g(t) + th(t) mid mathrm{d}t$$
?

Then I tried considering :
$$frac{G(h) - G(0)}{h}$$
In order to find the derivative. Yet the $mid cdot mid$ make the task not so easy.

So I tried the to split the integrand and study the part : $int_{Z(g)}$
I get (using the second interpretation of $G$) :

$$frac{int_{Z(g)} mid th(t) mid mathrm{d}t}{t} $$

But it doesn't seem to help...

It has to be
$$G(t)=int_0^1|g(x)+th(x)|,dx $$
otherwise $G(t)$ does not depend on $t$. We have
$$frac{G(t)-G(0)}{t}=int_0^1frac{|g(x)+th(x)|-|g(x)|}{t},dx= $$
By triangle inequality,
$$frac{|g(x)+th(x)|-|g(x)|}{t}leq |h(x)|in L^1(0,1) $$
Thus by dominated convergence, if $G'(0)$ exists, then
$$G'(0)=int_0^1lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t},dx$$
Now we distinsguish three cases.

• If $g(x)=0$, i.e. $xin Z_g$, then $$ lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0}frac{|t|h(x)}{t}$$
  

Notice that the above limit exists if and only if $h(x)=0$, and in this case, it is equal to $0$. Recall that we are assuming that $G'(0)$ exists. Now suppose by contradiction that there existed an $x_0in [0,1]$ such that $g(x_0)=0$ but $h(x_0)neq 0$. Since $Z_g$ is a finite union of intervals, we have $x_0in [a,b]$ with $g(x)=0$ on $[a,b]$. Moreover, since $h(x_0)neq 0$ and $h$ is continuous, we have $h(x)neq 0$ on $(x_0-varepsilon,x_0+varepsilon)subset [a,b]$. Thus the above limit does not exist on a whole interval $(x_0-varepsilon,x_0+varepsilon)$ (not just on $x_0$!). This implies that $G'(0)$ is not defined, contradicting our assumption. Therefore the above limit always exists and is equal to $0$.

• 
  

  If $g(x)>0$, then for $t$ small enough we have $g(x)+th(x)>0$. Thus

  
  
  

  $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{g(x)+th(x)-g(x)}{t}=h(x)=h(x)cdotoperatorname{sign}(g(x))$$

  
  


• 
  

  Similarly, if $g(x)<0$ then

  
  
  

  $$lim_{tto 0} frac{|g(x)+th(x)|-|g(x)|}{t}=lim_{tto 0} frac{-g(x)-th(x)+g(x)}{t}=-h(x)=h(x)cdotoperatorname{sign}(g(x))$$

  
  

Hence
$$G'(0)=int_{[0,1]setminus Z_g}h(x)cdot operatorname{sign}(g(x)),dx $$