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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. –  joriki Jun 13 '12 at 12:43
    
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. –  joriki Jun 13 '12 at 12:55
    
@joriki You can also write \| that is "backslash pine" –  AD. Jun 13 '12 at 13:15
    
for continuous function dont essential sup norm and sup norm coincide? –  clark Jun 13 '12 at 13:35
1  
@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. –  joriki Jun 13 '12 at 14:29

1 Answer 1

up vote 1 down vote accepted

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 } $

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thank you clark –  El Angel Exterminador Jun 13 '12 at 14:24

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