The discussion centers on proving the continuity of a linear operator T in a complex Banach space X, given the condition (T*f)(x)=f(Tx) for all x in X and f in X*. The participants conclude that the continuity of T can be established using the Uniform Boundedness Principle and the Hahn-Banach theorem. Specifically, they demonstrate that if the mapping x ↦ f(Tx) is continuous for each f in X*, then T must be continuous. The Closed Graph theorem is also mentioned as a viable alternative proof method.
PREREQUISITESMathematicians, particularly those specializing in functional analysis, graduate students studying operator theory, and anyone interested in the properties of linear operators in complex Banach spaces.
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 said: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?
Opalg said: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.
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.dray said: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.
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 $\displaystyle \sup_{x\in X_1}|S_x(f)| \leqslant \|T^*f\|.$dray said: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 said: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.
That is a very good idea. In fact, the closed graph theorem gives a simpler proof than the uniform boundedness method.dray said: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?