Here is Prob. 10, Sec. 3.2, in the book Introductory Functional Analysis With Applications by Erwin Kreyszig:
... Let $T \colon X \to X$ be a bounded linear operator on a complex inner product space $X$. If $\langle Tx, x \rangle = 0$ for all $x \in X$, show that $T=0$. ...
My Solution:
For any $x, y \in X$ and scalar $\alpha$, we have $$ \big\langle T(\alpha x + y), \alpha x + y \big\rangle = 0. \tag{1}$$ But using the linearity of $T$ and the properties of a complex inner product, we obtain $$ \begin{align} & \ \ \ \big\langle T(\alpha x + y), \alpha x + y \big\rangle \\ &= \big\langle \alpha Tx + Ty , \alpha x + y \big\rangle \qquad \mbox{[because $T$ is linear]} \\ &= \big\langle \alpha Tx, \alpha x \big\rangle + \big\langle \alpha Tx, y \big\rangle + \big\langle Ty, \alpha x \big\rangle + \big\langle Ty, y \big\rangle \\ &= \alpha \overline{\alpha} \langle Tx, x \rangle + \alpha \langle Tx, y \rangle + \overline{\alpha} \langle Ty, x \rangle + \langle Ty, y \rangle \\ &= \lvert \alpha \rvert^2 \langle Tx, x \rangle + \alpha \langle Tx, y \rangle + \overline{\alpha} \langle Ty, x \rangle + \langle Ty, y \rangle \\ &= \lvert \alpha \rvert^2 \cdot 0 + \alpha \langle Tx, y \rangle + \overline{\alpha} \langle Ty, x \rangle + 0 \\ & \qquad \qquad \mbox{[using the condition on $T$]} \\ &= \alpha \langle Tx, y \rangle + \overline{\alpha} \langle Ty, x \rangle. \tag{2} \end{align} $$ Using (1) and (2), we obtain $$ \alpha \langle Tx, y \rangle + \overline{\alpha} \langle Ty, x \rangle = 0. \tag{3} $$ In the last relation, putting $\alpha = 1$ and $\alpha = \iota$, respectively, we obtain $$ \langle Tx, y \rangle + \langle Ty, x \rangle = 0, \tag{4} $$ and $$ \iota \langle Tx, y \rangle - \iota \langle Ty, x \rangle = 0, $$ and this last equation upon dividing both sides by $\iota$ yields $$ \langle Tx, y \rangle - \langle Ty, x \rangle = 0, \tag{5} $$ Now adding (4) and (5), we get $$ \langle Tx, y \rangle = 0 \ \ \ \mbox{ for all } \ x, y \in X. \tag{6} $$ But for each $x \in X$, the image $Tx$ also is in $X$; so we can put $y = T(x)$ in (6) to obtain $$ \Vert Tx \Vert^2 = \langle Tx, Tx \rangle = 0 \ \ \ \mbox{ for all } \ x \in X, $$ and therefore $$ Tx = \mathbf{0} \ \ \ \mbox{ for all } \ x \in X, $$ Therefore we have shown that $T \colon X \to X$ is the zero operator, as required.
Is the above proof correct? If so, then where does the boundedness of $T$ come in?