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It is well known that the differential operator is an unbounded operator on the space of all continuously differentiable function on $[0,1]$. However,I found difficulties in finding an unbounded operator from $C[0,1]$ to $C[0,1]$, where $C[0,1]$ is the space of continuous function under sup-norm. Can someone explicitly give me an example of such operator?

EDIT: Can someone provide an unbounded operator from $X$ to $Y$ where $X$ and $Y$ are Banach space?($X,Y$ are to be determined)

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    You mean, everywhere defined and unbounded?2012-12-17
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    @abatkai: linear operator on the whole $C[0,1]$ but not bounded.2012-12-17
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    It's enough to find an unbounded linear functional, but I'm not sure we can give an explicit expression.2012-12-17
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    You can definitely give one by using a Hamel basis. This is not constructive, but quite good.2012-12-17
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    @Davide Giraudo: For the case linear functional,can you give me some hints?thank you.2012-12-17
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    As $C[0,1]$ is infinite dimensional, we can find a discontinuous linear functional $F$ using Hamel basis (it requires Zorn's lemma hence choice). Then define $T(f)(x):=F(f)$ to get a discontinuous linear operator. But I don't know whether an explicit formula of a discontinuous linear functional on $C[0,1]$ exists.2012-12-17
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    You have to use the axiom of choice to construct such an operator, so I would say this rules out having an "explicit" formula.2012-12-17
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    As Nate says, it is consistent that any linear functional between Banach spaces is continuous and therefore bounded in models where the axiom of choice fails. So one would have to either define the operator *from* an intangible object (e.g. a Hamel basis) which we cannot write explicitly either.2012-12-17
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    Since Banach space must have uncountable basis,is it true that it is not easy to (explicitly) find an unbounded operator between two arbitrary Banach spaces?2012-12-17

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