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Let $(C[0, 1], d_∞)$ be the metric space of continuous functions on $[0, 1]$ , where the distance function is defined by $$d_∞(f,g)= \sup_{x\in [0,1]}|f(x)-g(x)|$$

Consider the function $T:(C[0, 1], d_∞)\rightarrow (C[0, 1], d_∞)$ defined by

$$(Tf)(x):=\int_{0}^{x}f(t)dt$$

Why do we have $f(x)=e^x$ as a fixed point of $T$?

I am so confused because $T(e^x)=e^x-1$. But, $e^x \neq e^x-1$.

john
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    Where do you obtain the term -1? – ms_ Apr 27 '21 at 11:40
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    I just plugged $f(x)=e^x$ into $T$. Which gives us, $(Tf)(x):=\int_{0}^{x}e^tdt=e^x-1$ – john Apr 27 '21 at 11:42
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    @john Your question may be linked to my answer there: https://math.stackexchange.com/questions/4117556/proving-that-a-function-is-not-a-contraction/4117669#4117669 As the answer below shows, the constant in front of the exponential function must be zero, so there is indeed no fixed point. I have fixed my answer as well as it was not clear, and incorrect sometimes, thanks for your question. – Nicolas Apr 27 '21 at 11:46
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    @Nicolas Why do you say that there is no fixed point. Isn't the zero function a fixed point? – Kavi Rama Murthy Apr 27 '21 at 11:51
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    @KaviRamaMurthy Indeed, I meant "no non-trivial fixed point". By the way, I need to clarify it in my answer, thanks for your comment! – Nicolas Apr 27 '21 at 11:52

1 Answers1

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It is not. The only fixed point is $0$. $Tf=f$ implies $f'(x)=f(x)$ so $f(x)=ce^{x}$ for some constant $c$. But $Tf(0)=0$ so we must have $c=0$.

Kavi Rama Murthy
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