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Let $K: [0,1] \times [0,1] \to \mathbb{R}$ be continuous. Let $f_n : [0,1] \to \mathbb{R}^n$ be continuous and uniformly bounded. Define

$$g_n(x) = \int_{0}^{1} K(x,y) f_n(y) \ dy, \ \ x \in [0,1], n \geq 1.$$ Show that the sequence $\{g_n(x) \}$ is uniformly bounded and equicontinuous.

Attempt:

To show that the sequence is uniformly bounded, we must show there exists an $M$ such that $|g_n(x)| \leq M,\ \forall n \in \mathbb{N}$ and for all x.

Since $f_n(x)$ are uniformly bounded, we have that $|f_n(x)| \leq M'$. Since $K(x,y)$ is continuous on a compact interval, we have that $K(x,y)$ reaches a maximum := $N$. Thus, $$|g_n(x)| = \left|\int_{0}^{1} K(x,y) f_n(y) \ dy\right| \leq \left|M' \int_{0}^{1} K(x,y) \ dy\right| \leq |M'N(1-0)| := M.$$ Thus, the sequence $\{g_n(x)\}$ is uniformly bounded.

To show that the sequence is equicontinuous, we must show that for every $\epsilon > 0$ there exists a $\delta$ (which can depend on $x_0$) such that $|g_n(x) - g_n(x_0)| < \epsilon$ when $|x-x_0| < \delta.$ Well, $$|g_n(x) - g_n(x_0)| = \left|\int_{0}^{1} K(x,y) f_n(y) \ dy - \int_{0}^{1} K(x_0,y) f_n(y) \ dy\right| \leq M' \left|\int_{0}^{1} K(x,y) - K(x_0, y) \ dy \right|.$$

Since $K$ is uniformly continuous (because $K$ is continuous on a compact set), we have that $|K(x,y) - K(x_0, y)| < \epsilon/M'$ whenever $|x-x_0| < \delta$.

So, we get $$|g_n(x) - g_n(x_0)| \leq M' \left|\int_{0}^{1} K(x,y) - K(x_0, y) \ dy \right| \leq M' \int_{0}^{1} |K(x,y) - K(x_0, y)| \ dy \leq \ M'\epsilon/M' = \epsilon$$ whenever $|x-x_0| < \delta.$ Hence, the sequence is equicontinuous.

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    $\begingroup$ You dropped absolute values in the first part. And you have a mysterious $y_0$ in the second part. $\endgroup$ – zhw. Apr 17 '17 at 16:19
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Hint: Use the uniform continuity of $K$ on $[0,1]\times [0,1].$

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  • $\begingroup$ Ah, I think I get it. Does it look right now? $\endgroup$ – user389056 Apr 17 '17 at 17:26
  • $\begingroup$ That looks correct. You still have some housekeeping to do, like introduce $\epsilon$ earlier, tend to a few more absolute value signs. And is $[0,1]\times [0,1]$ an interval? $\endgroup$ – zhw. Apr 17 '17 at 17:32

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