Let $X$ and $Y$ be normed spaces, $T:X \rightarrow Y$ a linear operator. Show that $T$ is continuous if $y' \circ T$ is continuous for every $y' \in Y'$. My idea is the following: Suppose $T$ is not continuous, then there exists a sequence $x_n$ with $\|x_n\|=1$ such that $\|Tx_n\|>n$. If I could assure there is a subsequence $x_{n_k}$ such that $Tx_{n_k}$ are linearly independent I think I could construct an element $z'$ of $Y'$ such that $|(z' \circ T)x_{n_k}|>n_k$.

The first thing is, can I assure the existence of that subsequence? And if I can, how do I conclude applying H-B?

I'd like a solution with this same approach, but if there is a simpler proof i'd like to see it too.


This follows from the uniform boundedness principle. Consider the closed unit ball $B$ in $X$. Let $E=\{Tx:x\in X, \|x\|\leq1\}$. Then $y'(E)=(y' \circ T)(B)$ is bounded $\forall y' \in Y'$ by the continuity of $y' \circ T$. By the uniform boundedness principle, this implies that $E$ must be bounded, proving the continuity of $T$.


I have a proof using the closed graph theorem:

Let $y' \circ T$ be continuous for all $y' \in Y'$. Define the operator $T^* : Y' \to X'$ by $y' \mapsto y' \circ T$. Obviously, $T^*$ is linear. Moreover, it is not hard to show that $T^*$ is closed. Hence, it is bounded by the closed graph theorem, i.e., $\|T^*\| \le C$. This implies $$y'(T(x)) = (y'\circ T)(x) = (T^*y')(x) \le C \, \|y'\| \, \|x\|.$$ Hence, the norm of $T$ is bounded as well by $C$.

Note that $T^*$ is just the adjoint of $T$.

  • 1
    $\begingroup$ Hi gerw. Could you explain with a little more detail why $ \| T \| \leq C $? It doesn’t seem obvious to me because we only have $$ |y'(T(x))| \leq \| y' \| \cdot \| T(x) \| \quad \text{and} \quad |y'(T(x))| \leq C \| y' \| \cdot \| x \|. $$ There doesn’t seem to be a way to directly compare these two inequalities. $\endgroup$ – Berrick Caleb Fillmore Mar 1 '15 at 4:12
  • $\begingroup$ Using the last inequality and taking the sup over all $y'$ with norm at most $1$, we get $\|T(x)\| \le C \, \|x\|$. But this yields $\|T\| \le C$. $\endgroup$ – gerw Mar 4 '15 at 7:30

I wanted to make some comments on Arundhathi’s answer, which requires a little more attention to detail.

Let $ \mathbb{B} $ denote the closed unit ball of $ X $ and $ E \stackrel{\text{df}}{=} T[\mathbb{B}] $. Then for each $ \phi \in Y^{*} $, the set $$ \phi[E] = \phi[T[\mathbb{B}]] = (\phi \circ T)[\mathbb{B}] $$ is a bounded subset of $ \mathbb{C} $ by the assumption that $ \phi \circ T $ is continuous. Arundhathi brilliantly invokes the Uniform Boundedness Principle to conclude that $ E $ is a bounded subset of $ Y $, so $ T \in \mathscr{B}(X,Y) $.

For some people, the connection with the Uniform Boundedness Principle may not be very obvious. To flesh out this connection, let us first define a linear operator $ J: Y \to Y^{**} $ by $$ \forall y \in Y, ~ \forall \phi \in Y^{*}: \quad [J(y)](\phi) \stackrel{\text{df}}{=} \phi(y). $$ Applying the Hahn-Banach Theorem to the Banach space $ Y^{*} $ (equipped with the operator norm), it can be shown that $ J $ is an isometry.

Now, observe that for all $ \phi \in Y^{*} $, we have $$ \phi[E] = \{ [J(y)](\phi) \in \mathbb{C} \mid y \in E \}. $$ Hence, $ J[E] $ is a subset of $ Y^{**} $ that yields a bounded subset of $ \mathbb{C} $ upon evaluation at any $ \phi \in Y^{*} $. It then follows from the Uniform Boundedness Principle (its application being valid as $ Y^{*} $ is a Banach space, although $ Y $ may not be one itself) that $ J[E] $ is a bounded subset of $ Y^{**} $. As $ J $ is an isometry, we easily conclude that $ E $ is a bounded subset of $ Y $.

Upshot: The upshot of this discussion is that we have to be careful about the vector spaces that we are applying the Uniform Boundedness Principle to, because according to my limited knowledge, it does not hold between normed vector spaces in general. The same thing can be said for the Closed Graph Theorem.

For completeness, let me provide yet another solution using the Closed Graph Theorem. In order to be careful, we have to consider $ T $ as a linear operator from $ X $ to $ \bar{Y} $, the completion of $ Y $. This is not a problem however, because $$ T \in \mathscr{B}(X,Y) \quad \iff \quad T \in \mathscr{B} \! \left( X,\bar{Y} \right) ~ \text{and} ~ \text{Range}(T) \subseteq Y. $$

Let $ (x_{n})_{n \in \mathbb{N}} $ be a sequence in $ X $ such that $ (x_{n},T(x_{n}))_{n \in \mathbb{N}} $ converges to some $ (x,y) \in X \times \bar{Y} $.

On one hand, for each $ \phi \in Y^{*} $, we have \begin{align} \lim_{n \to \infty} (\phi \circ T)(x_{n}) & = \lim_{n \to \infty} \phi(T(x_{n})) \\ & = \bar{\phi} \! \left( \lim_{n \to \infty} T(x_{n}) \right) \qquad \left( \text{$ \bar{\phi} $ denotes the continuous extension of $ \phi $ to $ \bar{Y} $.} \right) \\ & = \bar{\phi}(y). \end{align} On the other hand, for each $ \phi \in Y^{*} $, we have \begin{align} \lim_{n \to \infty} (\phi \circ T)(x_{n}) & = (\phi \circ T) \! \left( \lim_{n \to \infty} x_{n} \right) \qquad (\text{As $ \phi \circ T $ is continuous.}) \\ & = (\phi \circ T)(x) \\ & = \phi(T(x)) \\ & = \bar{\phi}(T(x)). \end{align} If $ y \neq T(x) $, then by the Hahn-Banach Theorem applied to $ \bar{Y} $, there would be a $ \phi \in \bar{Y}^{*} $ separating $ y $ and $ T(x) $. By the argument above, this is impossible because any element of $ \bar{Y}^{*} $ is the continuous extension of some element of $ Y^{*} $. Hence, $ y = T(x) \in Y $, which yields $ (x,y) \in \text{Graph}(T) $.

Applying the Closed Graph Theorem to $ T: X \to \bar{Y} $, we conclude that $ T: X \to Y $ is continuous.

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    $\begingroup$ how can you apply Close graph until X and Y both are Banach sapaces? $\endgroup$ – Toeplitz Nov 1 '15 at 17:41

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