# Does this contradict the Closed Graph Theorem?

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.

-
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

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!

-
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