Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
The problem stems from the fact that (what I take to be) $(C^1[0,1],||\cdot||_{\infty})$ is not a Banach space. If you wanted it to be a Banach space you'd have to define a norm like $||f||_{C^1}:=||f||_{\infty}+||f'||_{\infty}$ In this case $T$ is bounded and there is no contradiction. If you consider the first normed space structure, the sequence $t^n$ is indeed bounded and has unbounded image through $T$, but in the Banach space structure, $t^n$ is unbounded... –  Olivier Bégassat Nov 1 '11 at 17:11
    
I changed "\quad for\quad each\quad" to "\text{ for each }". The former looks like this: ${}\quad for\quad each\quad{}$. The latter looks like this: $\text{ for each }$. –  Michael Hardy Nov 1 '11 at 18:21
    
@MichaelHardy: Thanks for the edit. –  Hassan Muhammad Nov 1 '11 at 19:44
add comment

1 Answer

up vote 10 down vote accepted

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!

share|improve this answer
5  
And under this new norm, $T$ would be bounded. –  Nate Eldredge Nov 1 '11 at 19:35
    
A variant of your sequence $x_n$ is also used in Pedersen's Analysis now for the proof of the Stone-Weierstrass theorem (see lemma 4.3.3). –  t.b. Nov 2 '11 at 15:53
    
@t.b.: thx for adding the link! –  math Nov 2 '11 at 19:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.