- Thread starter
- #1

- Thread starter dray
- Start date

- Thread starter
- #1

- Jan 30, 2012

- 2,533

But (T*f)(x)=f(Tx) is the definition of T*. I don't see how this imposes any constraint on T.

- Thread starter
- #3

- Jan 30, 2012

- 2,533

- Moderator
- #5

- Feb 7, 2012

- 2,793

As Evgeny.Makarov has pointed out, this question appears to be lacking some essential condition. My guess is that the question ought to be telling you that, for each f in X*, the map $x\mapsto f(Tx)$ is continuous. You could then use uniform boundedness to conclude that T is continuous.Let X be a complex Banach space and T in L(X,X) a linear operator. Assuming only that

(T*f)(x)=f(Tx), where x in X and f in X*

how can I prove that T is continuous?

- Thread starter
- #6

The question, stated in full is thus:As Evgeny.Makarov has pointed out, this question appears to be lacking some essential condition. My guess is that the question ought to be telling you that, for each f in X*, the map $x\mapsto f(Tx)$ is continuous. You could then use uniform boundedness to conclude that T is continuous.

(T*f)(x)=f(Tx), where x in X and f in X* (1)

defines a linear operator T* in L(X*,X*). Note that, whilst the domain and codomain of T* is the space of linear continuous functionals on X, you should notassume that T* is continuous.

Now, show that if we assume only equation (1), then that is sufficient to prove that T is continuous.

- Moderator
- #7

- Feb 7, 2012

- 2,793

That seems to agree with what I suggested. The missing information (which you have now supplied) is that T* belongs to L(X*,X*). In other words, for each f in X*, the map T*f is in X*, which means that it is aThe question, stated in full is thus:

Let X be a complex Banach space and T in L(X,X), a linear operator not assumed continuous. You may take it without proof that

(T*f)(x)=f(Tx), where x in X and f in X* (1)

defines a linear operator T* in L(X*,X*). Note that, whilst the domain and codomain of T* is the space of linear continuous functionals on X, you should notassume that T* is continuous.

Now, show that if we assume only equation (1), then that is sufficient to prove that T is continuous.

With that information, you can start to think in terms of uniform boundedness.

- Thread starter
- #8

- Moderator
- #9

- Feb 7, 2012

- 2,793

For each point $x$ in the unit ball $X_1$ of $X$, define a mapping $S_x:X^*\to\mathbb{C}$ (a linear functional on $X^*$) by $S_x(f) = T^*f(x)$. We are told that $T^*f$ is a continuous linear functional on $X$, and so $|T^*f(x)|\leqslant \|T^*f\|\|x\|$. Therefore $|S_x(f)| \leqslant \|T^*f\|\|x\|$, and $\displaystyle \sup_{x\in X_1}|S_x(f)| \leqslant \|T^*f\|.$I'm not that familiar with the Uniform Boundedness Principle, so wouldn't know where or how to apply this result to the question I'm afraid.

Thus $\displaystyle \sup_{x\in X_1}|S_x(f)|$ is bounded, with a bound $\|T^*f\|$ which on the face of it appears to depend on $f$. But the Uniform Boundedness theorem tells us that in fact this bound is uniform, in other words $\displaystyle \sup_{x\in X_1}\|S_x\| = K <\infty$. Hence $|f(Tx)| = |S_x(f)| \leqslant K\|f\|$ for all $f \in X^*$ and all $x\in X_1$. In particular, $\displaystyle \sup_{f\in X^*_1}|f(Tx)| \leqslant K$. It follows from the Hahn–Banach theorem that $\|Tx\|\leqslant K$ for all $x\in X_1$, in other words $T$ is bounded and therefore continuous.

- Thread starter
- #10

Thank you for replying to my question.For each point $x$ in the unit ball $X_1$ of $X$, define a mapping $S_x:X^*\to\mathbb{C}$ (a linear functional on $X^*$) by $S_x(f) = T^*f(x)$. We are told that $T^*f$ is a continuous linear functional on $X$, and so $|T^*f(x)|\leqslant \|T^*f\|\|x\|$. Therefore $|S_x(f)| \leqslant \|T^*f\|\|x\|$, and $\displaystyle \sup_{x\in X_1}|S_x(f)| \leqslant \|T^*f\|.$

Thus $\displaystyle \sup_{x\in X_1}|S_x(f)|$ is bounded, with a bound $\|T^*f\|$ which on the face of it appears to depend on $f$. But the Uniform Boundedness theorem tells us that in fact this bound is uniform, in other words $\displaystyle \sup_{x\in X_1}\|S_x\| = K <\infty$. Hence $|f(Tx)| = |S_x(f)| \leqslant K\|f\|$ for all $f \in X^*$ and all $x\in X_1$. In particular, $\displaystyle \sup_{f\in X^*_1}|f(Tx)| \leqslant K$. It follows from the Hahn–Banach theorem that $\|Tx\|\leqslant K$ for all $x\in X_1$, in other words $T$ is bounded and therefore continuous.

I had considered using the Closed Graph theorem but could only manage to show that (T*f)(x)=f(y). I'm not sure if you can even proceed from here to show that Tx=y and therefore deduce that T is continuous. Is it even possible?

Last edited by a moderator:

- Moderator
- #11

- Feb 7, 2012

- 2,793

That is a very good idea. In fact, the closed graph theorem gives a simpler proof than the uniform boundedness method.I had considered using the Closed Graph theorem but could only manage to show that (T*f)(x)=f(y). I'm not sure if you can even proceed from here to show that Tx=y and therefore deduce that T is continuous. Is it even possible?

Suppose that $x_n\to x$ and $Tx_n\to y$. The continuity of T*f shows (as you have noticed) that (T*f)(x)=f(y), for every f in X*. Therefore f(Tx)=f(y), or f(Tx-y)=0. Since that holds for all f in X*, it follows from the Hahn–Banach theorem that Tx=y. Thus T is closed and therefore continuous.

- Thread starter
- #12