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Problem: Show that $(C([a,b]),\|{\cdot}\|_2)$ is not a complete normed vector space.

I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question: Example of a non complete normed vector space. Although, I am running into difficulties and wondering if the example really generalises so well when we have a general $[a,b]$. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case?

Attempt: In order to show that $(C([a,b]),\|{\cdot}\|_2)$ is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in $C([a,b])$. In particular, it will suffice to show that our chosen Cauchy sequence converges under $\|\cdot\|_2$ to a discontinuous function.

Consider the following sequence of functions, $(f_n)_{n=1}^\infty\subset C([a,b])$, given by,

$$f_n(x)=\begin{cases} 0,\,\text{when}\,\,x\in[a,\frac{b-a}2-\frac{2^{-n}}{b-a}] \\[2ex] 1,\,\text{when}\,\,x\in[\frac{b-a}2+\frac{2^{-n}}{b-a},b] \\[2ex] \frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},\,\text{when}\,\,x\in[\frac{b-a}2-\frac{2^{-n}}{b-a},\frac{b-a}2+\frac{2^{-n}}{b-a}] \end{cases} $$

The definition of $f_n(x)$ when $x\in[\frac{b-a}2-\frac{2^{-n}}{b-a},\frac{b-a}2+\frac{2^{-n}}{b-a}]$ is the linear interpolant over that particular subinterval of $[a,b]$.

I claim that this is a Cauchy sequence. We consider:

$$\|f_n-f_m\|_2^2=\int_a^b|f_n(x)-f_m(x)|^2dx$$

$$=\int_a^{\frac{b-a}2-\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+\int_{\frac{b-a}2+\frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx$$

$$=0+\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0$$

The $0$'s for the first and last integrals because we have $f_n(x)-f_m(x)=0-0$ and $f_n(x)-f_m(x)=1-1$ on each respectively. As to the middle integral, here is what I've gotten so far:

$$\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx$$

Based in the example I am following, supposing that $m\ge n$, I think that I should be able to argue that the whole thing is bound above by $\frac{2^{-n}}{b-a}$ which would mean that $\|f_n-f_m\|_2^2\le\frac{2^{-n}}{b-a}\to0$ as $n\to\infty$.

I don't quite see how I'm meant to bound it above by $\frac{2^{-n}}{b-a}$, as I keep tripping up over applying that $m\ge n$, leading me to question whether the approach I have employed here is a fruitful one after all.

Masacroso
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Jeremy Jeffrey James
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1 Answers1

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Use the mean value theorem:

$$\int_a^b f(x)\,dx = (b-a)f(c), \text{ for some } c\in \langle a , b\rangle$$

In particular, since your $f_n(x) \in [0,1]$, we have:

\begin{align} \int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2\,dx &= \left(\frac{b-a}2+\frac{2^{-n}}{b-a} - \frac{b-a}2+\frac{2^{-n}}{b-a} \right)\underbrace{|f_n(c) - f_m(c)|^2}_{\le 2^2 = 4}\\ &\le 8\cdot \frac{2^{-n}}{b-a} \xrightarrow{n\to\infty} 0 \end{align}

Therefore, your sequence is Cauchy.

mechanodroid
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  • I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_n\to f$ as $n\to\infty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8\cdot \frac{2^{-n}}{b-a} \xrightarrow{n\to\infty} 0$ - is this correct? – Jeremy Jeffrey James Mar 15 '18 at 13:22