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PostPosted: Mon Jul 11, 2016 9:06 am 
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Consider the set \[ \displaystyle C \left( [a,b] \right) = \left\{ f : [a,b] \longrightarrow \mathbb{R} \; \Big| \; f : \text{continuous} \right\} \] Show that the function \[ \displaystyle d_{1} : C \left( [a,b] \right) \times C \left( [a,b] \right) \longrightarrow \mathbb{R} \; , \; d_{1} (f,g) = \int_{a}^{b} \big| f(t) - g(t) \big|\, \mathrm{d}t \] is a metric on \( \displaystyle C \left( [a,b] \right) \).
Afterwards, show that the metric space \[ \displaystyle \left( \; C \left( [a,b] \right) \, , \, d_{1} \; \right) \] is not complete.


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PostPosted: Mon Jul 11, 2016 10:20 am 
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Hi Nickos.

Let \(\displaystyle{f\,,g\,,h\in C(\left[a,b\right])}\).

Obviously, \(\displaystyle{d_1(f,g)\in\mathbb{R}\cap \left[0,+\infty\right)\,\,,\forall\,f\,,g\in C(\left[a,b\right])}\) and

\(\displaystyle{\begin{aligned} d_1(f,g)=0&\iff \int_{a}^{b}\left|f(t)-g(t)\right|\,\mathrm{d}t=0\\&\stackrel{|f-g|\geq 0}{\iff}\left|f-g\right|=0\\&\iff f=g\end{aligned}}\)

Also,

\(\displaystyle{d_1(f,g)=\int_{a}^{b}\left|f(t)-g(t)\right|\,\mathrm{d}t=\int_{a}^{b}\left|g(t)-f(t)\right|\,\mathrm{d}t=d_1(g,f)}\).

Finally,

\(\displaystyle{\begin{aligned} d_1(f,h)&=\int_{a}^{b}\left|f(t)-h(t)\right|\,\mathrm{d}t\\&=\int_{a}^{b}\left|(f(t)-g(t))+(g(t)-h(t))\right|\,\mathrm{d}t\\&\leq \int_{a}^{b}\left[\left|f(t)-g(t)\right|+\left|g(t)-h(t)\right|\right]\,\mathrm{d}t\\&=d_1(f,g)+d_1(g,h)\end{aligned}}\).

So, \(\displaystyle{d_1}\) is a metric on \(\displaystyle{C(\left[a,b\right])}\).

The metric space \(\displaystyle{\left(C(\left[a,b\right],d_1\right)}\) is not complete.

Indeed, consider \(\displaystyle{f_n:\left[a,b\right]\to \mathbb{R}\,,n\in\mathbb{N}\,,n\geq 3}\) by

\(\displaystyle{f_{n}(t)=\begin{cases}
0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,a\leq t\leq (a+b)/2\\
n\,\left(t-\dfrac{a+b}{2}\right)\,\,\,\,\,\,,(a+b)/2<t<(a+b)/2+(1/n)\\
1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,(a+b)/2+(1/n)\leq t\leq b
\end{cases}}\)

The sequence \(\displaystyle{\left(f_n\right)_{n\geq 3}}\) is a \(\displaystyle{\rm{Cauchy}}\) -sequence

since for \(\displaystyle{n>m}\) we get

\(\displaystyle{\begin{aligned} d_1(f_n,f_m)&=\int_{a}^{(a+b)/2}|f_n(t)-f_m(t)|dt+\int_{(a+b)/2}^{(a+b)/2+(1/m)}|f_n(t)-f_m(t)|dt+\int_{(a+b)/2+(1/m)}^{b}|f_n(t)-f_m(t)|dt\\&=\int_{(a+b)/2}^{(a+b)/2+(1/m)}|f_{n}(t)-f_{m}(t)|dt\\&\leq \dfrac{1}{m}\end{aligned}}\)

Suppose now that there exists \(\displaystyle{f\in C(\left[a,b\right])}\) such that \(\displaystyle{f_n\to f}\), that is

\(\displaystyle{\int_{a}^{b}|f_{n}(t)-f(t)|\mathrm{d}t\to 0}\). Then,

\(\displaystyle{0\leq \int_{a}^{(a+b)/2}|f(t)|\,\mathrm{d}=\int_{a}^{(a+b)/2}|f_n(t)-f(t)|\,\mathrm{d}t\leq \int_{a}^{b}|f_n(t)-f(t)|\,\mathrm{d}t}\)

and since \(\displaystyle{f\in C(\left[a,b\right])}\), we have that \(\displaystyle{f(t)=0\,,a\leq t\leq \dfrac{a+b}{2}}\).

Let \(\displaystyle{\delta\in\left(\dfrac{a+b}{2},b\right)}\). There exists \(\displaystyle{n_0\in\mathbb{N}}\)

such that \(\displaystyle{\dfrac{a+b}{2}+\dfrac{1}{n}<\delta\,,\forall\,n\geq n_0}\) and

\(\displaystyle{f_n(t)=1\,,t\in\left[\delta,b\right]}\). But,


\(\displaystyle{0\leq \int_{\delta}^{b}|f_{n}(t)-f(t)|dt\leq \int_{a}^{b}|f_{n}(t)-f(t)|dt\to 0}\) and

then \(\displaystyle{\int_{\delta}^{b}|1-f(t)|dt=0}\).

Since \(\displaystyle{f}\) is continuous, we get

\(\displaystyle{f(t)=1\,,\forall\,t\in\left[\delta,b\right]\,\,,\forall\,\delta\in\left(\dfrac{a+b}{2},b\right)}\)

and \(\displaystyle{f(t)=1\,,\forall\,t\in\left(\dfrac{a+b}{2},b\right]}\).

But now, \(\displaystyle{f}\) is not continuous at \(\displaystyle{t=\dfrac{a+b}{2}}\), a contradiction.

Therefore, \(\displaystyle{\left(C(\left[a,b\right]),d_1\right)}\) is not complete.


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