Proving $J$ is continuous by using contraction mapping theorem and proving that the sequence $J\beta_{n=1}^\infty$ has a convergent sub sequence

I am currently trying to solve and fill out the details of this problem:

Consider a continuous function $H : U = [0,1]\times [0,1] \to \mathbb R$. We define the norm of the function $H$ by $$\lVert H \rVert=\max_{(x,y)\in U}\lvert H(x,y) \rvert.$$ Now consider $\beta \in C([0,1], \mathbb R)$ and define the function $J \beta$ on $[0,1]$ by $J\beta(x) = \int_0^x H(x,y)\beta(y)dy$ where $x \in [0,1]$.

Questions ( and attempt at solution):

i) Show that the function $J\beta \in C[0,1]$ attempt:

so to do this we have to show that $J$ maps from $C[0,1] \rightarrow C[0,1]$. So in order to do so, I think we have to show the quantity the following quantity $J\beta(x+ \delta) - J\beta(x)$ is bounded ( any more insight on why this is true would be helpful) now this quantity is simply the integral $\int_0^{x+\delta} H(x+\delta, y)\beta(y)dy$ - $\int_0^xH(x,y)\beta(y)dy$.

Now i don't really know how to continue and manipulate this integral further. But I think since both H and $\beta$ are continuous on closed intervals then they are bounded and hence the integral bounded.

ii) Next we have to show that $J$ is continuous:

I think to do so we have to find a $C>0$ such that $\lVert J\beta \rVert$ $\leq C \lVert \beta \rVert$ for all $\beta \in C[0,1]$. I think we have to use the contraction mapping theorem. Since C[0,1] is complete then if we show that J is a contraction map we are done. This involves part (i) which is why I can't seem to figure out the details. We have to work out what C is in terms of $\lVert H \rVert$.

iii) Finally we have to show that if ${{\beta}}_{n=1}^\infty$ is a bounded sequence in $C[0,1]$ we have to show that the sequence $J\beta_{n=1}^\infty$ has a convergent subsequence.

This screams out Arzela-Ascoli theorem. So we have to show that $J\beta_{n=1}^\infty$ is bounded and that they are equicontinous. I'm not sure on how to do this.

Help would greatly be appreciated, as well as textbook recommendations or other external resources which might be helpful to this problem. Thank you in advance.

Since $U = [0,1]\times [0,1]$ is compact $H$ is uniformly continuous on $U$ so $$\forall \epsilon > 0,\, \exists \delta >0, \forall x,y\in [0,1],\, |x-y| < \delta \implies \forall s \in [0,1] ,\, |H(x,s) - H(y,s)| < \epsilon$$ or $$\forall \epsilon > 0,\, \exists \delta >0, \forall x,y\in [0,1],\, |x-y| < \delta \implies \, \sup_{s\in[0,1]}|H(x,s) - H(y,s)| < \epsilon$$ then you have for every $x, y\in[0,1]$ $$\sup_{s\in [0,1]} |H(x,s) - H(y,s)| \underset{x\to y} \to 0$$

i- Prove that $J\beta \in C([0,1], \mathbb R)$

For $x,y \in [0,1]$, $$|J\beta (x) - J\beta (y)| = \left|\int_0^x H(x,s)\beta (s)\mathrm ds - \int_0^y H(y,s)\beta (s)\mathrm ds\right| = \left|\int_0^x \left(H(x,s)-H(y,s)\right)\beta (s)\mathrm ds - \int_x^y H(y,s)\beta (s)\mathrm ds\right| \le \int_0^x |H(x,s) - H(y,s)||\beta(s)| \mathrm ds + \left|\int_x^y H(y,s) \beta (s)\mathrm ds\right| \le \left\|\beta\right\|\cdot\sup_{s\in[0,1]} |H(x,s) - H(y,s)| + \|H\|\|\beta\| |x-y| \underset{x\to y}\rightarrow 0$$

ii- Prove that $J : C([0,1], \mathbb R) \rightarrow C([0,1], \mathbb R)$ is continuous

We have only to prove that $\|J\beta\| \le C\|\beta\|$, which is true since $$|J\beta(x)| =\left|\int_0^x H(x,s)\beta (s) \mathrm d s\right| \le \int_0^x |H(x,s)||\beta (s)| \mathrm ds \le \|H\|\|\beta\| x \le \left\|H\right\| \left\|\beta\right\|$$

iii- If $(\beta_n)_n$ is bounded by $M$ prove that $(J\beta_n)_n$ has a convergent subsequence.

As you told this can be proven using Arzela-Ascoli theorem, let us prove that $(J\beta_n)_n$ is equicontinuous, indeed using the inequality proved in the first question : $$|J\beta_n (x) - J\beta_n (y)| \le \left\|\beta_n\right\|\cdot\sup_{s\in[0,1]} |H(x,s) - H(y,s)| + \|H\|\|\beta_n\| |x-y|\le M \left (\sup_{s\in[0,1]} |H(x,s) - H(y,s)| + \|H\||x-y|\right)$$ Now if you take $\epsilon > 0$ using the first remark you obtain $\eta > 0$ such that $$|x-y| < \eta \implies \sup_{s\in[0,1]} |H(x,s) - H(y,s)| + \|H\||x-y| < \epsilon$$ and you have the equicontinuity you were looking for.

• on part (ii) could you provide some more detail ? so is our constant C in this case just $\lVert H \rVert$ ? Also for part (iii) we get that the sequence $J\beta_n$ is bounded from part (i) and (ii) a bit more detail would be incredibly helpful. Thank you for your help! – user554863 Apr 23 '18 at 21:22
• Yes, indeed.... – Youem Apr 23 '18 at 21:23
• I hope it is clear now – Youem Apr 23 '18 at 21:25
• Could you please precise what is not clear for you ? – Youem Apr 23 '18 at 21:26