# Cauchy Sequence in $C[0,1]$ endowed with $L^2$ norm and $L^\infty$ norm

by ofir_13   Last Updated June 13, 2019 12:20 PM

Take $$V=C[0,1]$$ with the usual $$L^2$$ norm, i.e., $$\|f\|_2^2 = \int_{0}^{1}|f(\tau)|^2d\tau$$. Is it complete?

Consider the following sequence of functions:

$$f_n(t) = \begin{cases} 0 & t < \frac{1}{2} - \frac{1}{n+1} \\ [t-(\frac{1}{2}- \frac{1}{n+1})](n+1) & \frac{1}{2} - \frac{1}{n+1}\leq t \leq \frac{1}{2} \\ 1 & t \geq \frac{1}{2} \end{cases}$$ The sequence is Cauchy since $$\|f_m-f_n\|_2^2 = \frac{1}{m+1}$$, but obviously $$f_0 = \lim_{n \rightarrow \infty}f_n \notin C[0,1]$$.

Now do the same with the usual $$L^\infty$$ norm, i.e., $$\|f\|_\infty = sup_{t \in [0,1]}|f(t)|$$. I know that this space is complete, therefore I expect that the sequence $$f_n$$ is not a Cauchy sequence with the norm $$\|\cdot\|_\infty$$.

Let $$m \leq n$$, then:

$$\|f_n-f_m\|_\infty = sup_{t \in [\frac{1}{2} - \frac{1}{m+1},\frac{1}{2}]}\big(\frac{1}{2}-t\big)\big(n-m\big).$$

The question is: ''Is this a Cauchy sequence?''

I think no, because using the $$\epsilon$$ definition one can argue by contradiction such as: Assume $$\forall \epsilon > 0 \; \exists \; N_\epsilon > 0$$ s.t. $$\|f_n-f_m\|_\infty < \epsilon$$. Fix $$m = \hat{m} > N_\epsilon$$, then $$sup_{t \in [\frac{1}{2} - \frac{1}{m+1},\frac{1}{2}]}\big(\frac{1}{2}-t\big)\big(n-m\big)$$ depends only on $$n$$ and call $$\delta = \big(\frac{1}{2} -t\big)$$. Now take $$n > \frac{\delta\hat{m}+\epsilon}{\delta}$$, it follows that: $$\|f_n-f_m\|_\infty > \epsilon.$$

Is the reasoning correct?

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