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Let $X$ and $Y$ be real normed vector spaces and let $A: X \to Y$ be a bounded linear operator. Then we have $\text{im(A)}^\perp = \text{ker}(A^\ast)$.

The proof starts by saying let $y^\ast \in Y^\ast$. Then \begin{align} y^\ast \in \text{im(A)}^\perp & \Longleftrightarrow \langle y^\ast, Ax\rangle = 0 \ \text{for all} \ x \in X \\ & \Longleftrightarrow \langle A^\ast y^\ast, x\rangle = 0 \ \text{for all} \ x \in X \\ & \Longleftrightarrow A^\ast y^\ast = 0 \end{align}

and hence $y^\ast \in \text{ker}(A^\ast)$.

I don't see how we can go from $y^\ast \in Y^\ast$ to $y^\ast \in \text{im(A)}^\perp$. $y^\ast$ is a bounded linear functional from $Y$ to $\mathbb{R}$ so it acts on vectors/functions in $Y$, it's not actually in $Y$ itself which is what $y^\ast \in \text{im(A)}^\perp \subset Y$ implies.

So how can $y^\ast$ be in both $Y^\ast$ and $\text{im}(A)^\perp \subset Y$?

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    For a subset $K$ of $Y$ you define $K^\perp$ to be $K^\perp:=\{ y^*\in Y^*\mid y^*(k)=0 \,\forall k\in K\}$. So $\mathrm{im}(A)^\perp$ is a subset of $Y^*$. In the setting of a normed space (ie not an inner product space) there is no meaningful way to assign $\mathrm{im}(A)^\perp$ to subspace of $Y$. – 2017-01-23
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    Can you please explain why we are able to switch from $(y^*,Ax)$ to $(A^*y^*,x)$? Why $A$ is self-adjoint? – 2017-01-23
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    That has nothing to do with selfadjointness, $\langle y^*,Ax\rangle$ is a notation for $y^*(Ax)$ which is by definition the same as $A^*y^*(x)$ which in the other notation again becomes $\langle A^*y^*, x\rangle$ – 2017-01-23

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