Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 3 down vote accepted
  • $(1)$ 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.

share|cite|improve this answer
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. – user38355 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.

share|cite|improve this answer

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 $(4)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.