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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