TITLE: Internal and orthogonal direct sum isomorphic in Hilbert spaces
QUESTION [0 upvotes]: Let $H$ be a Hilbert space and let $E$ and $F$ be two closed subspaces of $H$ such that $H = E \oplus F$ as an internal direct sum. Can we conclude that $H$ is isomorphic to $E \times F$, where $E \times F$ is considered as a Hilbert space with the inner product $$\langle (x_1, y_1), (x_2, y_2) \rangle = \langle x_1, y_1 \rangle_H + \langle x_2, y_2 \rangle_H $$ via the map $i:x \mapsto (e_x, f_x)$, where $x = e_x + f_x$ is the decomposition in the internal direct sum?

If $F = E^\perp$, then this question is trivial.
However, we know that $F$ need not be equal to $E^\perp$ (Does orthogonal decomposition characterize direct sums in Hilbert space?). In this case, are we able to conclude they are isomorphic as Hilbert spaces via the canonical map? I do not know how to approach this.
I cannot conclude that $i$ is bounded directly, since $$||i(x)|| = \sqrt{||e_x||^2 + ||f_x||^2}, $$ which is not equal to $||e_x + f_x|| = x$, and it is always larger than this (by the triangle inequality).
Is it important that $H$ is a Hilbert space? We can ask the same question if $H$ is a Banach space. Will the answer change?

REPLY [1 votes]: It is a consequence of the closed graph Theorem, hence a non-trivial fact, that, whenever a Banach space $X$ is the (algebraic) direct sum of two closed subspaces $E$ and $F$, then the natural projections $\pi_E$ and $\pi_F$ are continuous.  Therefore, for every $x$ in $E$, the new norm
$$
   |||x||| = \|\pi_E(x)\| + \|\pi_F(x)\|
   $$
is equivalent to the original norm on $X$, and hence $X$ is isomorphic and homeomorphic to $E\times F$ via the map described by the OP.
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