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Prove that there is a unique bounded continuous real valued function $f:[0, \infty) \rightarrow \mathbb{R}$ such that $$ f(s)=1+\int_{0}^{s} e^{-t^{2}} f(s t) d t $$ for all $s \in[0, \infty)$

First as far as I searched $C[0,\infty)\cap B[0,\infty) $ is complete under sup metric.

So I tried to use Banach Fixed Point Theorem

similar to here https://math.stackexchange.com/a/1350747/342943

But I was confused about using the bound, which is not similar in the above question. $e^{-t^2}$ is going to overcome $f(t)$ for sufficiently bigger $n$.

Can you give me hint?

Or is there another approach that I can show above $f$ is unique and existent.

Gary
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Jale'de jaled
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1 Answers1

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The Banach fixed-point theorem can be used. As pointed out in the comments to On the solution of a Volterra type of integral equation, $T: C[0,\infty) \to C[0,\infty)$ defined by $$ Tf(s)= 1+ \int_{0}^s e^{-t^2} f(st)dt $$ maps bounded functions to bounded functions: $$ |Tf(s)| \le 1 + \int_{0}^s e^{-t^2} \, dt \cdot \Vert f \Vert_\infty \\ \le 1 + \int_{0}^\infty e^{-t^2} \, dt \cdot \Vert f \Vert_\infty = 1 + \frac{\sqrt \pi}{2 } \Vert f \Vert_\infty $$ and is a contraction: $$ |Tf(s) - Tg(s)| \le \int_0^s e^{-t^2} \, dt \cdot \Vert f-g \Vert_\infty \\ \le \int_0^\infty e^{-t^2} \, dt \cdot \Vert f-g \Vert_\infty = \frac{\sqrt \pi}{2 }\cdot \Vert f-g \Vert_\infty $$ because $\frac{\sqrt \pi}{2 } \approx 0.886 < 1$.

For the value of $\int_0^\infty e^{-t^2} \, dt $ see for example Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}$.

Martin R
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