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Let $K(t,s)$ be a continuous function on $[0,1]\times{[0,1]}$. Let $X=C[0,1]$ be the set of continuous functions defined on the interval $[0,1]$. Define the mapping $T:X\rightarrow{X}$ by:

for every $x\in{X}$, $T(x)(t)=\int^{1}_{0}K(t,s)x(s)ds$ for all $t\in{[0,1]}$.

(i) Let $d$ be the supremum metric on $X$. Show that $T$ is continuous on $(X,d)$.

(ii) Let $d_{2}$ be the $L_{2}$ metric, i.e., $d_{2}(x,y)=\sqrt{\int^{1}_{0}[x(s)-y(s)]^{2}ds}$ for $x,y\in{X}$. Show that T is continuous on $(X,d_{2})$.

Sorry in advance but I need a lot of help on this.

For (i), do I need to introduce a $t_{0}$ such that $t_{0}\in{[0,1]}$? Then $T(x)(t)-T(x)(t_{o})=\int^{1}_{0}K(t,s)x(s)ds-\int^{1}_{0}K(t_{0},s)x(s)ds$.

mmh0015
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1 Answers1

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Hints and clarification:

(i) To show that $T$ is continuous on $(X,d)$ means to show that for any function $f\in X=C[0,1]$ and any $\epsilon>0$, there exists $\delta>0$ such that if $g\in X$ and $d(f,g)<\delta$, then $d(T(f),T(g))<\epsilon$. In particular, do NOT introduce a $t_0$. Here $d$ is your supremum metric!

(ii) Is the same, with a different metric.

Gyu Eun Lee
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  • Thank you. I understand the definition but don't know how to use it. Obviously let $\epsilon>0$ and let $f\in{X}=C[0,1]$. – mmh0015 Mar 07 '14 at 02:20
  • Can I use neighborhoods and say... Let $\epsilon>0$ and let $f\in{X}=C[0,1]$. Then $f^{-1}(O(T(g),\epsilon))$ is open in $X$. So there exists $\delta>0$ such that $O(g,\delta)\subseteq{f^{-1}(O(T(g),\epsilon))}$ which implies $T(O(g,\delta))\subseteq{O(T(g),\epsilon)}$. Thus, $T$ is continuous on $(X,d)$. – mmh0015 Mar 07 '14 at 02:36
  • No. $f^{-1}(O(T(g),\epsilon)$ doesn't even make any sense, $f$ actually doesn't have much to do with this problem. The problem is about $T$, think of $f$ as just a point in a metric space. This is actually quite a simple problem, why don't you write down exactly what $d(T(f),T(g))$ means and try to find an upper bound for it? – Gyu Eun Lee Mar 07 '14 at 03:30
  • I'm assuming $d(T(f),T(g))=\int^{1}{0}K(t,s)|f(s)-g(s)|ds\leq{\int^{1}{0}K(t,s)d(f,g)ds}\leq{\int^{1}_{0}K(t,s)ds\cdot{d(f,g)}}$? – mmh0015 Mar 07 '14 at 03:55
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    Yes, good (except I don't know where the $q$ came from, but doesn't matter). Now, since $K$ is continuous the integral is just a constant. Can you choose $\delta>0$ so that $d(f,g)<\delta$ makes this $<\epsilon$? – Gyu Eun Lee Mar 07 '14 at 03:57
  • I'm not sure where it came from either. Since the integral of $K$ is just a constant, say $c$, then I can write the above equal to $c\cdot{d(f,g)}$. Maybe choose $\delta=\frac{\epsilon}{c}$? Then $d(T(f),T(g))=c\cdot{d(f,g)}<{c\cdot{\delta}}\leq{c\cdot{\frac{\epsilon}{c}}}= \epsilon$. – mmh0015 Mar 07 '14 at 04:06
  • Yes, precisely. You can do a similar argument for (ii). Not so hard, right? – Gyu Eun Lee Mar 07 '14 at 04:13
  • Not at all! Thank you so, so much for your help (: – mmh0015 Mar 07 '14 at 04:14
  • Quick question for part (ii). The metric is different, but wouldn't it still be the same as part (i) in that choose $\delta>0$ such that $\delta=\frac{\epsilon}{c}$. Then $d_{2}(T(f), T(g))=\int^{1}{0} K(t,s)\sqrt{\int^{1}{0}[|f(s)-g(s)|]^{2}}ds\leq{\int^{1}{0} K(t,s)d{2}(f,g)ds}\leq{\int^{1}{0} K(t,s)ds\cdot{d{2}(f,g)}}<c\cdot{d_{2}(f,g)}\leq{c\cdot{\delta}}\leq{\frac{\epsilon}{c}}={\epsilon}$? – mmh0015 Mar 12 '14 at 23:55
  • This argument fails: what you have there in the first equality is NOT the definition of $d_2(T(f),T(g))$. As a hint, I suggest the Cauchy-Schwarz inequality. – Gyu Eun Lee Mar 13 '14 at 04:55