If $A$ and $A^\perp$ are closed then $A\oplus A^\perp=B$ for a Banach space $B$

Question

Let $b$ be a bounded, positive definite and symmetric bilinear form on a Banach space $B$.

I want to prove the following:

Let $A\subset B$ be a closed subspace such that $A^\perp$ is closed too. Then $B=A\oplus A^\perp$. $(A^\perp=\{x\in B\vert b(x,a)=0\,\forall a\in A\})$

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Clearly $A\cap A^\perp=0$ since $b$ is positive definite. But how can I prove that $A+A^\perp = B$? Would it help if i knew that $A + A^\perp$ is closed?

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Edit: Its wrong. Counter example in the answer and here: When does $A^⊥=0$ imply that $A=X$ for a subspace $A⊂X$ of a Banach space

Answer 1

I believe this is false. Consider the Banach space $B = (\ell^2,\|\cdot\|_2)$ and let $$b(x,y) = \sum_{n=1}^\infty \frac1n x_ny_n, \qquad x=(x_n)_{n=1}^\infty, y = (y_n)_{n=1}^\infty \in \ell^2.$$ Let $$A = \left\{y = (y_n)_{n=1}^\infty \in \ell^2 : \sum_{n=1}^\infty \frac1ny_n = 0\right\}.$$ Then $A$ is a closed subspace and $A^\perp = \{0\}$ is also a closed subspace but $A \dotplus A^\perp = A \subsetneq \ell^2$.