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I'm studying for an exam in real analysis. Thus, only such techniques should be considered. I'm looking at old exams, and repeatedly see questions similar to the one below.

Show that there exists a unique non-zero solution $f(x)$ to the integral equation  $$f(x)=\frac{1}{2}\int_0^1\sin(xy)f(y)dy.$$ Here, $C([0,1])$ denotes the space of continuous functions $\phi$ on the interval $[0,1]$ with the norm $\|\phi\|:=\sup_{x\in[0,1]}|\phi(x)|$.

I assume that the solution lays in the properties of the Riemann-Stieltjes integral, but I'm unsure about what techniques to use. The course literature is Walter Rudin's book on real analysis, which doesn't mention integral equations at all.

The other problems often only give bounds for the kernel such as $|K(x,y)|\leq\frac{1}{2}$.

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$$\left|\frac12\int_0^1\sin(xy)f(y)\mathrm{d}y\right|\leq\frac12\int_0^1\left| \sin(xy)f(y)\right|\mathrm{d}y\leq\int_0^1 |f(y)|\mathrm{d}y\leq\frac12||f||_\infty,$$

hence the map $f(x)\mapsto\frac12\int\dots$ is a contraction of $\mathcal C([0,1])$, so it has a unique fixed point by the Banach fixed point theorem

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  • $\begingroup$ Why is the fixed point a solution? $\endgroup$ – Logarithmic Derivative Aug 8 '18 at 10:24
  • $\begingroup$ A fixed point has the property:$f(x)=\frac{1}{2}\int_0^1\sin(xy)f(y)dy.$ Since $f=0$ is a fixed point, the unique solution of the integral equation is $f=0$. $\endgroup$ – Fred Aug 8 '18 at 11:05
  • $\begingroup$ Why does it have that property? In general, a fixed point is only a point $c$ such that $f(c)=c$ for some function $f$, right? $\endgroup$ – Logarithmic Derivative Aug 8 '18 at 11:13
  • $\begingroup$ Define $T:C[0,1] \to C[0,1]$ by $(Tf)(x):= \frac{1}{2}\int_0^1\sin(xy)f(y)dy.$. Then $f$ is a solution of the integral equation $ \iff Tf=f \iff f$ is a fixed point of $T$. $\endgroup$ – Fred Aug 8 '18 at 11:15
  • $\begingroup$ Okay. Thanks, Fred. $\endgroup$ – Logarithmic Derivative Aug 8 '18 at 11:19
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  1. $f=0$ is a solution.

  2. If $f(x)=\frac{1}{2}\int_0^1\sin(xy)f(y)dy$ for $x \in [0,1]$, then

    $|f(x)| \le \frac{1}{2}\int_0^1|\sin(xy)| \cdot |f(y)| dy \le \frac{1}{2}\int_0^1||f|| dy= \frac{1}{2} ||f|| $

for all $x \in [0,1]$.

This gives $||f|| \le \frac{1}{2} ||f|| $ , hence $f=0.$

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  • $\begingroup$ The solution has to be non-zero. $\endgroup$ – Logarithmic Derivative Aug 8 '18 at 9:46
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    $\begingroup$ 2. in my answer shows: if $f$ is asolution of the integral equation, then we have $f=0$. Consequence: the integral equation has no non-zero solution. $\endgroup$ – Fred Aug 8 '18 at 11:03
  • $\begingroup$ I highly doubt that an exam question would entail showing something that isn't true. However, I do see your argument. $\endgroup$ – Logarithmic Derivative Aug 8 '18 at 11:11

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