# Why is such an operator continuous?

These two questions were in one question of a list of exercises.

Let $E$ be a Banach space and $T : E \longrightarrow E^*$ be linear.

• If $\langle T(x),x \rangle \geq 0$ holds for all $x \in E$, then $T$ is continuous.

• If $\langle T(x),y \rangle = \langle x ,T(y) \rangle$ holds for all $x, y \in E$, then $T$ is continuous.

I tried to expand it, as in the proof of the Cauchy-Schwarz inequality, to get a polynomial of degree $2$. Any solution or hint?

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You can use \langle and \rangle to get the inner product ($\langle ... \rangle$) brackets –  Tyler Aug 24 '13 at 21:37
By $\langle T(x),x\rangle$ do we mean $(T(x))(x)$? Edit: no, that can't be it. What, though? –  Jonathan Y. Aug 24 '13 at 21:41
@JonathanY. Yes –  user40276 Aug 24 '13 at 21:41
How can you switch the roles, then? Are we implicitly using the embedding of $E$ in $(E^*)^*$? –  Jonathan Y. Aug 24 '13 at 21:43
@JonathanY It's actually fairly common notation to denote the pairing $E^\ast \times E \to \mathbb{C}$ of a Banach space $E$ with its dual $E^\ast$ just like an inner product, e.g., $\langle \cdot,\cdot \rangle$, so that $$\forall f \in E^\ast, e \in E, \quad \langle f,e \rangle := f(e).$$ In this case, indeed, $\langle T(x),y\rangle := T(x)(y)$. The notation $\langle x, T(y) \rangle$, likewise, has to refer to the dual pairing $E^{\ast\ast} \times E^\ast \to \mathbb{C}$ of $E^\ast$ with its dual $E^{\ast\ast}$, with the canonical injection $E \to E^{\ast\ast}$ used implicitly. –  Branimir Ćaćić Aug 24 '13 at 22:06

• $(1)$ is true also in the real case. Here is one possible proof although I'm not sure its the quickest (since its an adaption of a proof showing (possibly nonlinear) monotone operators are locally bounded). If $T$ were not bounded, then there would exist a sequence $x_n \to 0$ with $\|Tx_n\| \to \infty$. Define $$c_n = 1 + \|Tx_n\|\|x_n\|.$$ Now let $z \in E$. Then by assumption $$0 \le \langle T(z - x_n), z - x_n \rangle$$ which after expanding and rearranging turns into $$\langle Tx_n, z \rangle \le \langle Tx_n, x_n - z \rangle + \langle Tz, z - x_n \rangle.$$ Since $c_n > 1$, we get $$c_n^{-1}\langle Tx_n, z \rangle \le c_n^{-1}\langle Tx_n, x_n - z \rangle + \langle Tz, z - x_n \rangle$$ $$\le 1 + c_n^{-1}\|Tz\|\|z - x_n\| \le M(z)$$ where $M(z)$ is some constant independent of $n$. We can repeat the same argument with $-z$ in place of $z$ to get $$-c_n^{-1}\langle Tx_n, z \rangle \le M(-z)$$ where again $M(-z)$ is independent of $n$. Thus we can use the Banach-Steinhaus Theorem to conclude that $$\sup c_n^{-1}\|Tx_n\| \le C < \infty.$$ Recalling the definition of $c_n$ we get $$\|Tx_n\| \le C(1 + \|Tx_n\|)\|x_n\|$$ so $$(1 - C\|x_n\|)\|Tx_n\| \le C$$ for all $n$. This implies $\|Tx_n\| \le 2C$ when $\|x_n\| \le \frac{1}{2C}$ contradicting the fact that $\|Tx_n\| \to \infty$ as $x_n \to 0$. So $T$ is bounded.

• Here is also an alternative to $(2)$ which mimicks the Hellinger-Toeplitz Theorem. Let $x_n \to x$ in $E$ be such that there exists $y \in E^*$ with $Tx_n \to y$. Then we have $$\langle y, z \rangle = \lim \langle Tx_n,z \rangle = \lim \langle Tz, x_n \rangle$$ $$= \langle Tz, x \rangle = \langle Tx, z \rangle$$ for all $z \in E$ (where we used continuity of the linear functional $Tz$ in the third equality). This means that $y = Tx$ and therefore the graph of $T$ is closed. Hence $T$ is continuous by the Closed Graph Theorem.

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Why Hahn-Banach? If functionals coincide at each point, then they´re equal. –  user40276 Aug 27 '13 at 13:02
@user40276 Yeah, right. I was confusing what I was proving with the dual statement that $f(x) = f(y)$ for all $f \in E'$ implies $x = y$. I've edited the post. –  brom Aug 27 '13 at 13:48

For $x\in S_{E}$ consider the family $F_x$ of bounded functionals given by $$F_x(y)=\langle Tx,y\rangle$$

We have $|F_x(y)|=|\langle Tx,y\rangle|=|\langle x,Ty\rangle|\leq||x||\cdot||Ty||=||Ty||$

Therefore, the family $\{F_x : x\in S_E\}$ is pointwise bounded, and it follows from the Uniform Boundness Principle that is also norm bounded. Note that this works for both $T$ symmetric and anti-symmetric.

Since the family is norm bounded, there exists $K$ such that for any $x\in S_E$, we have $$||Tx||=\sup_{y\in S_E}|\langle y, Tx\rangle| <K$$

which means that $T$ is bounded, thus showing $(2)$.

For $(1)$, if the vector spaces are over $\mathbb{C}$, the condition implies the anti-symmetry of $T$, as Daniel Fischer noted before. The proof above works just as well for $T$ anti-symmetric. Don't know about the real case.

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Here is an alternative proof for the first statement.

Suppose that $x_n\to x$ in $E$ and $Tx_n\to f$ in $E'$. By hypothesis, we have that $$\langle Tx_n-Ty,x_n-y\rangle\ge 0,\ \forall \ y\in E\tag{1}$$

If we pass the limit in $(1)$ we get that $$\langle f-Ty,x-y\rangle\ge 0,\ \forall\ y\in E\tag{2}$$

Now take $y=x+tv$ where $t\in \mathbb{R}$ and $v\in X$. We have that $$\langle f-Tx-tTv,-tv\rangle\geq 0,\ \forall\ t\in\mathbb{R},\ v\in E\tag{3}$$

We get from $(3)$ that $-t\langle f-Tx, v\rangle\geq-t^2\langle Tv,v\rangle$ or equivalently $$\langle f-Tx, v\rangle\leq t\langle Tv,v\rangle,\ \forall\ t> 0,\ v\in E \tag{4}$$

Now it is straightforward to conclude from $(4)$.

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