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Question:

Let C^{'}[0,1] and $C[0,1]$ be endowed with sup norm. Define T:C^{'}[0,1]\to C[0,1] by Tf=f^{'}\text{ for each }f\in C^{'}[0,1] Where, ', indicates differentiation.

Prove that $T$ is a linear map with closed graph but it is not bounded.


$Proof:$

For linearity:

Let $f_{1}$, f_{2}\in C^{'}[0,1] $\alpha,\beta$ scalars, then T(\alpha f_{1}+\beta f_{2})=(\alpha f_{1}+\beta f_{2})^{'}=\alpha f_{1}^{'}+\beta f_{2}^{'}=\alpha Tf_{1}+\beta Tf_{2}.

For the closer of $T$,

Let ${f_{n}}$ be a sequence in C^{'}[0,1] such that $f_{n}\to f$ and Tf_{n}=f_{n}^{'}\to y. Then

\|Tf_{n}-y\|=\sup_{t\in [0,1]}|T(f_{n})(t)-y(t)|=\sup_{t\in [0,1]}|T(f_{n}^{'})(t)-y(t)|\to 0 as $n\to \infty$

Thus, the convergence is uniform and y(t)=\lim_{n\to \infty}f_{n}^{'}(t). Since the convergence is uniform f^{'}(t)=y(t) for all $t\in [0,1]$. Thus, f\in C^{'}[0,1] and $Tf=y$ so $T$ is closed.

To show $T$ is not bounded, take $f_{n}(t)=t^{n}$, then $\|f_{n}\|=\sup_{t\in [0,1]}|t^{n}|=1$ and f^{'}_{n}(t)=nt^{n-1} so that $\|Tf_{n}\|=\sup_{t\in [0,1]}|nt^{n-1}|=n$.

Thus, $T$ is not bounded. This implies $T$ is not continuous.


My question here is, does this example contradict the Closed Graph theorem? Which says: If you have two Banach spaces $X$ and $Y$ and $T$ a linear map from $X$ to $Y$ such that the graph of $T$ is closed. Then $T$ is continuous.

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    @MichaelHardy: Thanks for the edit.2011-11-01

1 Answers 1

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No it's not, because the (general) space $ C^1[a,b] $ is not a Banach space (with this norm). Of course it is a subspace of $ C[a,b]$ but not closed (with respect to the sup norm. Therefore it is not a Banach space) To see that it's not closed, take for example:

$ a=-1, b=1$ and $x_n(t) = (t^2+\frac{1}{n})^\frac{1}{2}$

this is a sequence of continuously differentiable functions such that it converges uniformly to $ |\cdot|$, and this is not differentiable! This is an example from Dirk Werner's "Funktionalanalysis".

cheers

math

EDIT: if you take instead the norm $ \|x\|:=\|x\|_\infty + \|x^\prime\|_\infty $ you get a Banach space!

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    @t.b.: thx for adding the link!2011-11-02