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Problem:

Let $C[0,1]$ denote the collection of all real continuous functions defined on $I=[0,1]$.

Consider the metrics $d$ and $e$ on $C[0,1]$ defined by

$$d(f,g) = \sup\{|f(x) - g(x)|: x \in I\}$$

$$e(f,g) = \int_0^1 |f(x) -g(x)|\:\text{dx}$$

Show that the topology $T_d$ induced by metric $d$ is not coarser than the topology $T_e$ induced by metric $e$.


Solution:

I am supposed to show that $T_d \not \subset T_e$.

Let $p(x)$ be the constant function $p(x) = 2$ and let $\epsilon = 1$. Then the ball(s) $B_d(p, \epsilon)$ consists of all functions $g(x)$ for which $g(x)$ lies between the functions $p(x) - 1$ and $p(x) + 1$, i.e. such that $1 < g(x) < 3$ for all $x \in I$.

It is sufficient to show that $B_d(p, \epsilon)$ contains no $e$-open ball with center $p(x)$; i.e. for every $\delta > 0$ $B_e(p, \delta) \not \subset B_d(p, \epsilon)$.

I am not sure on how I should continue.

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Given $\delta >0$, we can define a function $g$ by $g(x) = \frac{-8}{\delta}x+4$ if $x \in [0,\frac{\delta}{4}]$ and $g(x)=2$ otherwise. Then $g(0)=4$ so that $g$ is not in the open ball $B_d(p,\varepsilon)$. Furthermore $$\int^1_0 |g(x)-p(x)|dx = \int_0^{\frac{\delta}{4}} \frac{-8}{\delta}x+4-2 \ dx = \frac{-4}{\delta}x^2 +2x|_0^{\frac{\delta}{4}} = -\frac{\delta}{4}+\frac{\delta}{2} = \frac{\delta}{4} < \delta$$

Therefore $g \in B_e(p, \delta)$ for all $\delta >0$.

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  • $\begingroup$ Thanks, i will look into your answer in detail tomorrow. $\endgroup$ – JKnecht Mar 31 '16 at 23:45
  • $\begingroup$ Pleasure, note that $\int^1_0 |g-f| dx$ is just the area of the triangle with vertices $(0,2)$, $(0,4)$ and $(0,\frac{\delta}{4})$. $\endgroup$ – sqtrat Mar 31 '16 at 23:47
  • $\begingroup$ The idea is that e(f,g) can be close to 0 even though f(x)-g(x) can be very large for some x, just as an area (an integral) of a very long thin shape can be very small . $\endgroup$ – DanielWainfleet Mar 31 '16 at 23:55
  • $\begingroup$ @sqtrat I was thinking along the same lines as you when i was trying to solve this on my own. What confuses me is $1.$ Yes, we can find some $g$ where $g(x) \not \in B_d(p,\varepsilon)$. But $2.$ We can also find another function $h(x)$ with $h(x) \in B_e(p,\delta) \subset B_d(p,\varepsilon)$. It feels like we randomly pick a $g(x)$ that fits our need. I know im wrong here and that i am misunderstanding something. Are you able to explain what is wrong in my reasoning? $\endgroup$ – JKnecht Apr 1 '16 at 20:53

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