Linear Operator

dray

Member
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?

Evgeny.Makarov

Well-known member
MHB Math Scholar
But (T*f)(x)=f(Tx) is the definition of T*. I don't see how this imposes any constraint on T.

dray

Member
Perhaps I should restate the final remark and say "how is this sufficient to show that T is continuous".

Evgeny.Makarov

Well-known member
MHB Math Scholar
I don't think it suffices. Indeed, take any operator T and defined T* on L* by the equation (T*f)(x) = f(Tx). I don't see how defining T* tells us anything about T. Operator T does not even have to be linear to define T*.

Opalg

MHB Oldtimer
Staff member
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?
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.

dray

Member
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.
The 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.

Perhaps Hahn Banach and Closed Graph theorem?

Opalg

MHB Oldtimer
Staff member
The 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.
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 a continuous linear functional on X.

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

dray

Member
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.

Opalg

MHB Oldtimer
Staff member
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.
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.

dray

Member
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.
Thank you for replying to my question.

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?

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Opalg

MHB Oldtimer
Staff member
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?
That is a very good idea. In fact, the closed graph theorem gives a simpler proof than the uniform boundedness method.

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.

dray

Member
Aha!

I never thought to use Hahn Banach.

Thanks