# Why does this show that $T$ is contractive on $Y$?

Consider the Cauchy problem $$\dot{x}=f(x,t),~~~x(t_0)=x_0,$$ where $f$ and $f_x'$ are continuous. Now we proved that there is a unique solution on $[t_0-\delta,t_0+\delta]$. To do so, we defined the operator $$T(x(t)):=x_0+\int_{t_0}^t f(x(s),s)\, ds$$ and showed that it is (i) invariant and (ii) contractive on $Y$, where $Y$ contains all continuous functions on $[t_0-\delta,t_0+\delta]$. The statement then follows from Banach contraction principle.

I did not understand the proof that $T$ is contractive:

For this, we proved that $$\lVert\Delta T(x(t))\rVert\leq q\cdot\sup_{\lvert s-t_0\rvert\leq\delta}\lVert\Delta x(s)\rVert$$ for some $q<1$.

In the lecture, we wrote that $$\Delta T(x(t))=T(x(s)+\Delta x(s))-T(x(s)).$$

But why does this show that $T$ is contractive on $Y$?

(I only know the definition of being contractive as $$\lVert T(x(s))-T(y(s))\rVert\leq q \lVert x(s)-y(s)\rVert$$ for some $q<1$.)

• You did not define $\Delta T$ and $\Delta x$. – Siminore Oct 23 '15 at 14:32
• It was not defined in the lecture neither. But we wrote that $\Delta T(x(t))=T(x(s)+\Delta x(s))-T(x(s))$. – M. Meyer Oct 23 '15 at 14:33

$Y$ is a normed space with norm defined as $\|x\|=\sup_{|s-t_0|\le\delta}|x(s)|$. From here we get a distance $d(x,y)=\|x-y\|$. What you have shown is $$\|\Delta T(x)\|=_{\text{def}}\|T(x+\Delta x)-T( x)\|\le q\|\Delta x\|=_{\text{def}}q\|(x+\Delta x)-x\|.$$ Call $y=x+\Delta x$. The above inequality is now $$\|T(y)-T(x)\le q\|y-x\|.$$ This is the definition of a contraction in $Y$.

• But the supremum in my inequality is irritating me. – M. Meyer Oct 23 '15 at 14:42
• The supremum over $s$ is precisely the definition of the norm in $Y$. You may be confusing $|\cdot|$ and $\|\cdot\|$. You should write $$\sup_s|\Delta x(s)|=\|\Delta x\|$$. – Julián Aguirre Oct 23 '15 at 14:46
• Ah great, I will write the responsible person because of his confusion of $\lvert\cdot\rvert$ and $\lVert\cdot\rVert$. – M. Meyer Oct 23 '15 at 14:50

We suppose $t_0\in [a,b]$ and $f,f'_x$ continuous on $[a,b]$

$\|T(x(s)-T(y(s))\| =\|\int_{t_0-\delta}^{t_0+\delta}(f(x(s),s)-f((y(s),s))ds \|$.

Since $f'_x$ is continuous, $\| f(x(s),s)-f((y(s),s)\|\leq$ $Sup \|f'_x\|_{x\in [a,b]}\|x(s)-y(s)\|$.

$|T(x(s)-T(y(s))\|\leq 2\delta Sup \|f'_x\|_{x\in [a,b]}\|x(s)-y(s)\|$. We can choose $\delta$ such that $2\delta Sup \|f'_x\|_{x\in [a,b]}=q<1$. This implies that $\|T(x(s)-T(y(s))\|\leq q\| x(s)-y(s)\|$