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Let (X, ||.||) be a Banach space and T: X $\rightarrow$ $X^*$ a linear operator with the property (Tx)(y) = (Ty)(x) for every x, y $\in$ X.

Let $(X, ||\cdot||)$ be a Banach space and $T: X \rightarrow X^*$ a linear operator with the property $(Tx)(y) = (Ty)(x)$ for every $x, y \in X$. To prove that $T$ is a bounded operator, we need to show that $T$ maps bounded sets in $X$ to bounded sets in $X^*$.

Let $B$ be a bounded set in $X$, which means there exists a constant $M > 0$ such that $||x|| \leq M$ for all $x \in B$.

Now, consider the set $\{Tx : x \in B\}$ in $X^*$. We want to show that this set is bounded in $X^*$.

For any $x \in B$ and $y \in X$, by the property of $T$ given, we have:

$|(Tx)(y)| = |(Ty)(x)| \leq ||Ty|| \cdot ||x||$

Have I started correct? How to continue?

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2 Answers 2

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You want to use the Closed Graph theorem. This is usually one of the best tools to use when you want to prove that a linear map between Banach spaces is continuous.

Since $X$ is Baanch we have that $X^{*}$ is too. Now, let $(x_n,Tx_n) \longrightarrow (x,\varphi) \in X \times X^{*}$. It suffices to prove that $Tx = \varphi$ to prove that the graph of $T$ is closed.

Indeed, observe for $y \in X$, we have: $$(Tx)y = (Ty)x = \lim_n (Ty)(x_n) = \lim_n (Tx_n)y = \varphi(y)$$ where the second equality follows from $Ty$ being a continuous linear functional on $X$. Hence since $y \in X$ was arbitrary, we have $Tx=\varphi$ so that $T$ is bounded by the Closed Graph theorem.

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As an alternative to using the closed graph theorem, you can use the uniform boundedness principle.

We want to show that $\sup_{\|x\|= 1} \|T x\| < \infty$. By the uniform boundedness principle applied to the set of functionals $\{T x \in X^* : \|x\| = 1\}$, it suffices to show that for each fixed $y \in X$, $\sup_{\|x\| \le 1} |T(x)(y)| < \infty$.

But this supremum can be written as $\sup_{\|x\|\le 1} |T(y)(x)| = \|Ty\| < \infty$ since $y$ is fixed $Ty \in X^*$

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