# Functional analysis - bounded linear transformation

Let $\mathcal{H}$ be a Hilbert space, and let $T: \mathcal{H} \to \mathcal{H}$ be such that $\langle x,Ty \rangle = \langle Tx,y \rangle$ for all $x,y \in \mathcal{H}$.

How can one show that $T$ is linear and bounded? It would be great if someone could use the Closed Graph Theorem to prove the result. Thanks!

-
Formatting problems render this unintelligible. – ncmathsadist Jan 5 '13 at 2:47
-1: You've been here long enough to know that you don't ask homework questions by copying them from your textbook. – Nate Eldredge Jan 5 '13 at 5:36
Apparently two people thought "This question shows research effort. It is useful and clear." Seriously? – Martin Jan 5 '13 at 6:05

We first prove that $T$ is linear. Let $\mathbf{x}_{1},\mathbf{x}_{2} \in \mathcal{H}$ and $\lambda \in \mathbb{F}$, where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$. Then \begin{align} \forall \mathbf{y} \in \mathcal{H}: \quad \langle T(\mathbf{x}_{1} + \lambda \cdot \mathbf{x}_{2}),\mathbf{y} \rangle &= \langle \mathbf{x}_{1} + \lambda \cdot \mathbf{x}_{2},T(\mathbf{y}) \rangle \\ &= \langle \mathbf{x}_{1},T(\mathbf{y}) \rangle + \langle \lambda \cdot \mathbf{x}_{2},T(\mathbf{y}) \rangle \\ &= \langle \mathbf{x}_{1},T(\mathbf{y}) \rangle + \lambda \langle \mathbf{x}_{2},T(\mathbf{y}) \rangle \\ &= \langle T(\mathbf{x}_{1}),\mathbf{y} \rangle + \lambda \langle T(\mathbf{x}_{2}),\mathbf{y} \rangle \\ &= \langle T(\mathbf{x}_{1}),\mathbf{y} \rangle + \langle \lambda \cdot T(\mathbf{x}_{2}),\mathbf{y} \rangle \\ &= \langle T(\mathbf{x}_{1}) + \lambda \cdot T(\mathbf{x}_{2}),\mathbf{y} \rangle. \end{align} Hence, $$\forall \mathbf{y} \in \mathcal{H}: \quad \langle T(\mathbf{x}_{1} + \lambda \cdot \mathbf{x}_{2}) - [T(\mathbf{x}_{1}) + \lambda \cdot T(\mathbf{x}_{2})],\mathbf{y} \rangle = 0.$$ By choosing $\mathbf{y} = T(\mathbf{x}_{1} + \lambda \cdot \mathbf{x}_{2}) - [T(\mathbf{x}_{1}) + \lambda \cdot T(\mathbf{x}_{2})]$, we see that \begin{align} T(\mathbf{x}_{1} + \lambda \cdot \mathbf{x}_{2}) - [T(\mathbf{x}_{1}) + \lambda \cdot T(\mathbf{x}_{2})] &= \mathbf{0}, \quad \text{or equivalently}, \\ T(\mathbf{x}_{1} + \lambda \cdot \mathbf{x}_{2}) &= T(\mathbf{x}_{1}) + \lambda \cdot T(\mathbf{x}_{2}). \end{align} As $\mathbf{x}_{1},\mathbf{x}_{2},\lambda$ are arbitrary, we conclude that $T$ is a linear operator.
We now prove the continuity of $T$. Let $(\mathbf{x}_{n})_{n \in \mathbb{N}}$ be a sequence in $\mathcal{H}$ that converges to $\mathbf{0}$, and suppose that $\displaystyle \lim_{n \to \infty} T(\mathbf{x}_{n}) = \mathbf{y}$. By the Closed Graph Theorem, it suffices to show that $\mathbf{y} = \mathbf{0}$. We proceed as follows. \begin{align} 0 &= \langle \mathbf{0},T(\mathbf{y}) \rangle \\ &= \left\langle \lim_{n \to \infty} \mathbf{x}_{n},T(\mathbf{y}) \right\rangle \\ &= \lim_{n \to \infty} \langle \mathbf{x}_{n},T(\mathbf{y}) \rangle \\ &= \lim_{n \to \infty} \langle T(\mathbf{x}_{n}),\mathbf{y} \rangle \\ &= \left\langle \lim_{n \to \infty} T(\mathbf{x}_{n}),\mathbf{y} \right\rangle \\ &= \langle \mathbf{y},\mathbf{y} \rangle. \end{align} Therefore, $\mathbf{y} = \mathbf{0}$ as required.