I was wondering if $C([0,1])$ is a complete space in all $p$. I know that $$\| \cdot \|_1 \leq \| \cdot \|_{\infty},$$ but am not sure if it holds for all $p$.
$C([0,1])$ complete in all $p$ norms
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functional-analysis
norm
cauchy-sequences
complete-spaces
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1To complement the answer given below: Note that the inequality given is not sufficient to guarantee completeness. We need to have an inequality in the other direction aswell. – 2017-02-18
1 Answers
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Define $$ f_j(x) := \begin{cases} 0 &\text{ if } x < \frac12 - \frac1j \\ \frac12 + \frac j2(x -\frac12) &\text{ if } \left|x - \frac12\right| \leq \frac1j \\ 1 &\text{ if } x > \frac12 + \frac1j \end{cases} $$ Then $f_j \in \mathrm C([0,1])$. If $f := \mathbf 1_{[1/2,1]}$, we have $f_j \to f$ in measure, and $|f_j| \leq 1$, so $f_j \to f$ in $\mathrm L^{\!p}$. Meanwhile, $f$ is not continuous, so this shows a sequence in $\mathrm C([0,1])$ may converge in $\mathrm L^{\!p}$ to a discontinuous function.
If $p = \infty$ your statement indeed holds, since for any continuous function, $\lVert f \rVert_{\mathsf u} = \lVert f \rVert_\infty$.