Prove that the equation $\phi(x)=\int_{0}^{x}\frac{(x-t)^{n-1}}{(n-1)! }\phi(t) dt $ have trivial solution for each $n\in \mathbb{N}$

Question

Prove that the equation $$\phi(x)=\int_{0}^{x}\frac{(x-t)^{n-1}}{(n-1)! }\phi(t) dt $$ have trivial solution for each $n\in \mathbb{N}$ in $C[0,\alpha]$

Proof

I want prove that exists a transformation $T:C[0,\alpha]\to C[0,\alpha]$ such that $T^{n}$ is contractil for conclude that $T$ have a unique fixed point say $x_0$ and conclude that $\phi(x_0)=x_0=0$.

For it I assume that $T_{n}(\phi(t))=\int_{0}^{x}\phi_{n-1}(t)dt$ and hence $T_{n}^{n}=\phi(x)$.

From here I need find $\gamma\in(0,1)$ such that $d(T(x)-T(y))\leq \gamma d(x,y)$.

Let $\phi_1(t),\phi_2(t)$arbitrary functions in $C[0,\alpha]$.

$$Sup_{x\in [0,\alpha]}=\left|\int_{0}^{x}\frac{(x-t)^{n-1}}{(n-1)! }\phi_1(t) dt -\int_{0}^{x}\frac{(x-t)^{n-1}}{(n-1)! }\phi_2(t) dt \right|=Sup_{x\in[0,\alpha]}\frac{1}{(n-1)!}\left| \int_{0}^{x} (x-t)^{n-1}(\phi_1(t)-\phi_2(t))dt\right|$$ From here I need bound it by $Sup_{x\in[0,\alpha]} |\phi_1(t)-\phi_2(t)|$.

Since $0\leq x\leq \alpha$ then $0 \leq x^n \leq \alpha^n$ and hence

from here I can´t continue but I think that $\gamma$ should be $\gamma=\frac{1}{(n-1)!}$.

Any advice for this proof?

Answer 1

I think you are misinterpreting the problem, we want to show that the solution is trivial for each $n\in \mathbb{N}$.

Fix $n\in\mathbb{N}$ and consider $\varphi_1,\varphi_2\in C[0,\alpha]$ and define $$T_n\varphi=\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}\varphi(t)dt$$ Note I will use $\|\cdot \|_\infty$ to denote the sup-norm on $[0,\alpha]$ and consider $$\|T_n\varphi_1-T_n\varphi_2\|_\infty=\Big\|\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}(\varphi_1(t)-\varphi_2(t))dt \Big\|_\infty$$ since $\varphi_1,\varphi_2\in C[0,\alpha]$ we know they each obtain their supremums and you can move the absolute value inside the integral to get $$\|T_n\varphi_1-T_n\varphi_2\|_\infty\leq\Big\|\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}|\varphi_1(t)-\varphi_2(t)|dt \Big\|_\infty$$ $$\leq \|\varphi_1(t)-\varphi_2(t)\|_\infty \Big\|\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}dt\Big\|_\infty$$

Now we compute $\Big\|\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}dt\Big\|_\infty$ directly $$\Big\|\int_0^x\frac{(x-t)^{n-1}}{(n-1)!}dt\Big\|_\infty=\Big\| \frac{x^n}{n!}\Big\|_\infty=\frac{\alpha^n}{n!}$$ Now, for every $n>1$ this says that $T_n$ is a contractive map and since the zero function satisfies the equation it must be the only solution.

Consider the case $n=1$ on your own.