Suppose $T: L^2(\mathbb R^d) \to L^2(\mathbb R^d)$ is a linear operator, and there exists $T^*: L^2(\mathbb R^d) \to L^2(\mathbb R^d)$ such that $\langle Tx, y \rangle = \langle x, T^*y \rangle$ for all $x, y \in \mathbb R^d$. Prove that $T$ is bounded.
I have no idea how to approach this problem. I think it says that $T$ is "invertible" if it is bounded, which makes sense. I have no idea how to prove it though. Right now I'm a bit confused by the inner product. We haven't seen anything in class about how to treat a linear operator in an inner product, so I assume that the reason we are restricted to $L^2$ is that it is self-dual. I don't see how that helps at all though. Any help at all will be appreciated.
EDIT: So I managed to solve the other 5 questions from this same homework without any major issues, but I still have no idea how do solve this one. The homework has already been handed in, but I want to understand this question. My lecturer is away at a conference, so it will be a week and a half before I can ask him about it. Everyone else I've asked just tells me to apply the closed graph theorem, just like Topoguy says in his answer. The problem is that I don't understand how to show that the graph is closed. I assume it's something really obvious that I'm missing. Could someone explain to me why everyone seems to think it's really obvious to apply the closed graph theorem to this problem?
It's equivalent to prove $T$ is continuous. One theorem to get $T$ is continuous is the closed graph theorem. So can you show the graph of ($x$,$Tx$) is closed? Suppose $x_n \rightarrow x$ and $Tx_n \rightarrow y$ and we want to show $Tx=y$. Indded, $\langle Tx, z \rangle = \langle x, T^*z \rangle =\lim_{n \to \infty} \langle x_n, T^*z \rangle = \lim_{n \to \infty}\langle Tx_n, z \rangle= \langle y, z \rangle $. Therefore, $\langle Tx-y,z\rangle=0$ for all $z\in L^2$ and hence $Tx=y$ which means the graph is closed, i.e. $T$ is continuous.