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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.

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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.

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