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would any one tell me whether $C[0,1]$ is complete under these metrics

1.sup norm i mean $\|f\|_{\infty}$

2.$\|f\|_{\infty,1/2}=\|f\|_{\infty}+|f(1/2)|$

3.$\|f\|_{2}=\sqrt{\int_0^1|f|^2dx}$

Under supnorm I know it is complete,I am not sure about the other two.

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    a) You can get proper formatting for the norm symbols by using `\lVert\cdot\rVert` to produce $\lVert\cdot\rVert$. b) The unusual notation in part $3$ is unnecessarily confusing; $\lVert f\rVert_2$ is usually defined to be the square root of what you wrote.2012-06-13
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    In fact what you wrote isn't a norm, since it doesn't satisfy the triangle inequality. Note also that you're not distinguishing clearly between norms and metrics.2012-06-13
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    @joriki You can also write \| that is "backslash pine"2012-06-13
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    for continuous function dont essential sup norm and sup norm coincide?2012-06-13
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    @clark Sorry, I'd already deleted that comment because I realized it was misguided, as you point out.2012-06-13
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    @Mex: For part $3$, assuming you mean the usual $\|\cdot\|_2$ norm with the square root included, consider the sequence of functions $f_n=1/(1+\exp(n(x-1/2)))$, and try to show that this is a Cauchy sequence without a limit in $C[0,1]$. Or take any similar sequence of functions, e.g. functions that are $0$ and $1$ in the left and right half of the interval, respectively, except for a linear ramp whose width decreases with $n$.2012-06-13
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    @joirki, how did you find that? amazing...2012-06-13
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    @Mex: My thought process was something like this: First I considered an example where the functions have a narrowing peak of constant height and vanish everywhere else. The problem is that such a sequence does converge to the zero function with respect to the metric induced by $\lVert\cdot\rVert_2$. So it had to be something that also has a narrowing structure, but where a discontinuity remains in the limit. That suggested having two different values amd a transition between them instead of a peak.2012-06-13
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    beautiful arguement and thought process. thanx again2012-06-13
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    @Mex: By the way, if errors in the question are pointed out in comments (such as the missing square root), please correct them to make the question correct and self-contained so people don't have to read through the comments to understand it.2012-06-13
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    @AD: The advantage of `\lVert` and `\rVert` over `\|` is that they lead to the right spacing. For instance, `\lVert\cdot\rVert` produces $\lVert\cdot\rVert$, whereas `\|\cdot\|` produces $\|\cdot\|$.2012-06-13
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    @joriki Ok, I see the difference. Knowing this I think it is a matter of taste.2012-06-13

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for the second case the norm you asked is equivalent to the sup norm so the normed space it is induced also complete $ |\!|f|\!| _{\infty} \leq |\!| f |\!|_{\infty , \frac{1}{2}} =|\!|f|\!| _{\infty}+ |f(\frac{1}{2}) | \leq 2|\!|f|\!| _{\infty } $