# Adjoint of a surjective bounded linear operator on a Hilbert Space

Let $$H$$ be a Hilbert Space and let $$T \in BL(H, K)$$ be surjective then prove that the adjoint of $$T$$ is bounded below.

Since $$T$$ is surjective, so for any $$y \in K, \exists$$ $$x \in H$$ such that $$Tx = y$$ $$\implies ||y||=||Tx|| \leq ||T||.||x|| \implies ||y|| \leq c||x||$$, for some $$c>0$$.

We also define a unique $$T^*: K \to H$$ such that $$\langle Tx, y\rangle = xT^*y$$, as the adjoint of $$T$$.

It would be very helpful if I could get an insight of how to proceed from here. Thanks.

Suppose that $$T^*$$ is not bounded below. There exists a sequence $$(f_n\in H^*)$$ such that $$\|f_n\|=1$$ and $$\|T^*(f_n) \|<1/n$$. This implies that
$$sup_{\|x\|=1}|T(f_n(x))|<1/n$$.
Since $$T$$ is surjective open mapping theorem implies that $$T(B_K(0,1))$$ contains $$B_H(0,s)$$. Since $$\|f_n\|=1$$, There exists $$y_n\in H$$ such that $$\|y_n\|=1$$ and $$|f_n(y_n)|\geq 1/2$$. Since $$T$$ is surjective, $$sy_n=T(x_n)$$. We can suppose that $$x_n\in B_K(0,1)$$. This implies that $$|f_n(T(x_n)|=|T^*(f_n) (x_n)|\geq s/2$$. We deduce that $$\|T^*(f_n) \|\geq s/2$$ for every $$n$$. Contradiction.