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My question is, given a vector $x$ in a normed space, and any two subspaces with an intersection $0$ and whose direct sum is the whole vector space, is the norm of projection along one of the subspaces of the vector $x$ always less than or equal to the norm of $x$ itself? Thanks for your help.

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  • $\begingroup$ What about if you replace "projection along" with closest point in the subspace (when it $\exists!$)? Hmm... $\endgroup$ – Pp.. Jan 17 '15 at 4:29
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No. Consider the real plane, and the subspaces $y = 0$ and $y = x/10$. The projection of $v = (0, 1)$ along the first subspace (onto the second) is $(10, 1)$.

On the other hand, if the two subspaces are orthogonal, then the projections along each space, onto the other, are indeed shorted than the original vector.

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No, this isn't even true in general for norms induced by inner products: Consider $\mathbb{R}^2$, and the decomposition $\mathbb{R}^2 = X \oplus Y$ into the $x$- and $y$-axes, so that the projections of $(x, y)$ onto $X$ and $Y$ are respectively $(x, 0)$ and $(0, y)$.

Now, consider the inner product given in the standard basis by $$\langle (x, y), (x', y') \rangle = A x x' + B(x y' + x' y) + C y y';$$ this is an inner product iff $AC - B^2 > 0$ and (e.g.) $A > 0$, and the induced norm is $$N((x, y)) = \sqrt{\langle (x, y), (x, y) \rangle} = \sqrt{A x^2 + 2 B x y + C y^2}.$$

The vector $(1, 1)$ has norm $A + 2B + C$ and its projection $(1, 0)$ has norm $A$, so we can produce a counterexample just by finding a triple $(A, B, C)$ such that $$AC - B^2 > 0 \qquad \text{and} \qquad A + 2B + C < A,$$ and this is satisfied, for example, by $A = 2, B = -1, C = 1$, giving the norm $$N((x, y)) = \sqrt{2 x^2 - 2 x y + y^2}.$$

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