There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$

Asked Jun 25 '13 at 01:20

Active Jun 25 '13 at 01:20

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I'm doing past papers for a first course on functional analysis. We are not allowed to assume any results from real analysis or topology, so I was surprised to find an exam question, where I couldn't say "follows from version 13 of Hahn-Banach":

Let $\sin(x-y)$ be defined on $[0,1]\times[0,1]$ and $A$ be a real constant such that $|A|<1$. Show that for any $f\in C([0,1])$, there exists a unique $g\in C([0,1])$ such that

$$g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$$

I have absolutely no idea how even to start. I tried integrating by parts (twice) and got nowhere. Could someone give me a hint?

edited Jun 12 '20 at 10:38

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asked Jun 25 '13 at 01:20

Earthliŋ

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related problems:(I), (II). – Mhenni Benghorbal Jun 25 '13 at 01:24
By the way, if you benefit from these answers, consider up voting them. – Mhenni Benghorbal Jun 25 '13 at 01:34
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@MhenniBenghorbal Cheers, let me have a closer look. – Earthliŋ Jun 25 '13 at 01:37
You are welcome. – Mhenni Benghorbal Jun 25 '13 at 01:39
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@MhenniBenghorbal I like your answer here. Successive iteration gives existence (provided one checks that the sequence is Cauchy). Any suggestions for uniqueness? – Earthliŋ Jun 25 '13 at 02:12
Another related problem. – Mhenni Benghorbal Jun 25 '13 at 02:18
Note that, if the limit exists then it is unique. – Mhenni Benghorbal Jun 25 '13 at 02:34
@MhenniBenghorbal I have a Cauchy sequence converging to a unique limit, which satisfies the property. I do not know, whether there are other functions which satisfy this property, though. The unique limit doesn't show uniqueness of the function satisfying the property. – Earthliŋ Jun 25 '13 at 02:37
It converges in a complete space. Note that, you have a contraction operator and if you assume that you have two fixed points, then you will have a contradiction. – Mhenni Benghorbal Jun 25 '13 at 02:55
I'm not even sure how to define the contraction operator without assuming that $g$ is unique, though. – Earthliŋ Jun 25 '13 at 03:21
Note that you are given that $|A|<1$. – Mhenni Benghorbal Jun 25 '13 at 03:53
@MhenniBenghorbal I understand that $T\colon f\mapsto g$ is a contraction mapping, if it is well-defined, which is only more or less obvious if $g$ is unique in the first place. – Earthliŋ Jun 25 '13 at 03:55

There exists unique $g$ s.t. $g(x) = f(x) + A\int_0^1\sin(x-y)g(y)dy$

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