does it have unique fixed point?

$p:C[0,1]\rightarrow C[0,1]$ defined by $p(f(x))=\int_{0}^{x} (x-t)f(t)dt$, well, I am getting all constant functions are fixed points, but the answer says that it has unique fixed point. I got $p(f(x))=f(x)$ so $f'(x)=0$.

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I am getting that the only constant function that is a fixed point is the zero constant function – Amr Jan 13 '13 at 11:42

Let $f$ be a fixed point. Then $x\int_0^xf(t)dt-\int_0^xtf(t)dt=f(x)$ so in particular $f$ is derivable. We have $$f'(x)=\int_0^xf(t)dt+xf(x)-xf(x)$$ so $f'(x)=\int_0^xf(t)dt$. Let $g(x):=\int_0^xf(t)dt$. We have $g''(x)=g(x)$, and if we don't forget the initial conditions, $g'(0)=g''(0)=0$, which gives an unique $g$ (namely, $0$) hence and unique fixed point $f(x)=0$ for all $x$.

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Hint: $$|p(f)(x)| \leq \|f\|_{C([0,1])}\int_0^x(x-t)\,dt$$ and the integral $\int_0^x (x-t)\,dt$ is bounded in $x$ by a constant strictly less than 1. Now you can estimate the $C([0,1])$-norm of $p(f)$ and then apply the Banach fixed-point theorem to get the existence of a unique fixed point.

Since $p$ is a linear operator, the constant function $0$ is obviously a fixed point, hence the unique one.

Edit: one doesn't even need the fixed-point theorem:

As above, $0$ is obviously a fixed point and the contraction property above gives for fixed points $f,g \in C([0,1])$: $$\|f-g\|_{C([0,1])} = \|p(f)-p(g)\|_{C([0,1])} = \|p(f-g)\|_{C([0,1])} \leq q \|f - g\|_{C([0,1])}$$ with $q < 1$, hence $\|f-g\|_{C([0,1])} = 0$ and thus $f = g$, i.e. fixed points are unique.

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