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Let's consider $L_2(\mathbb{R}^n)$. Let $Y$ be a non empty closed subspace of $L_2(\mathbb{R}^n)$.

Let $x\notin Y$. Let $y^*$ be the best approximation of $x$ on $Y$, i.e., $\|x-y^*\|_2=\inf_{y\in Y}\|x-y\|_2$.

We know then that, $x-y^*$ would be orthogonal to $Y$ and hence from parallelogram law, one can deduce the pythagoras theorem: $$\|x-y\|_2^2=\|x-y^*\|_2^2+\|y^*-y\|_2^2 \text{ for } y\in Y$$

I'm wondering whether the same kind of result would be true for $L_p(\mathbb{R}^n)$, $p\ge 1$, $p\neq 2$ also, i.e., whether $$\|x-y\|_p^p=\|x-y^*\|_p^p+\|y^*-y\|_p^p $$

I think its not possible to deduce from the parallelogram law as we have only inequality in parallelogram law in $L_p(\mathbb{R}^n)$ and there's no notion of orthogonality in $L_p(\mathbb{R}^n)$ for $p\neq 2$. But I think there may be some other way to get the result. At least mentioning some reference is appreciated.

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Do you mean that $Y$ is a non-empty closed subspace of $L_2(\mathbb{R}^n)$? (BTW, you don't need to specify convex, if it is an actual vector subspace). – Martin Wanvik Feb 8 '12 at 12:35
Yes. I will correct it. – Ashok Feb 8 '12 at 14:44
up vote 8 down vote accepted

The statement is false.

First note that $\mathbb{R}^n$ with the $p$-norm embeds in $\mathcal{L}^p(\mathbb{R}^n)$: just map the unit vectors to indicator functions of any $n$ disjoint sets of unit mass. Also, finite-dimensional subspaces are always closed. Thus a necessary condition for the statement to hold is that it holds for subspaces $Y$ of $\mathbb{R}^n$ with the $p$-norm.

Take $Y=\{(x,x),x\in\mathbb{R}\}\subset\mathbb{R}^2$, $x=(0,2)$, and $1<p<\infty$. For any $y\in Y$ define $\hat{y} = (2,2) - y$. Then $\lVert x-\hat{y}\rVert_p = \lVert x-y\rVert_p$. Thus $\frac{y+\hat{y}}{2}=(1,1)$ is the average of two points on the boundary of the $p$-ball of radius $\lVert x-y\rVert_p>0$ centered at $x$. All nontrivial $p$-balls are strictly convex, so $(1,1)$ is strictly closer to $x$ than $y$ is unless $y = \hat{y} = (1,1)$. Therefore $y^* = (1,1)$. At $y=0\in Y$ the equation you have written reduces to $\frac{4}{2^p}=1$, so it can only hold for $p=2$.

This example is somewhat problematic at $p=1$ because $p$-balls are not strictly convex and there is not a unique minimizer $y^*$. However, changing to $Y=\{(x,2x)\vert x\in\mathbb{R}\}$ gives a unique minimizer and the desired equation again fails to hold.

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First of all, thanks for the answer. Though the last two paragraphs answer my question, I don't understand the second paragraph fully. Do you mean, by $\mathcal{L}^p(\mathbb{R}^n)$, the space of all functions $f$ from $\mathbb{R}^n$ to $\mathbb{C}$ s.t. $\|f\|_p<\infty$? Why do we want to think of $L_p(\mathbb{R}^n)$ embedded in $\mathcal{L}^p(\mathbb{R}^n)$? Also, can you elaborate on "By symmetry and strict convexity of $p$-balls..."? – Ashok Feb 9 '12 at 5:31
@Ashok: You're welcome. I elaborated a bit on the convexity argument; please let me know if it suffices now. As for the second paragraph, I think I may have misread your original question. Perhaps you were intending to ask about $\mathbb{R}^n$ with the $p$-norm all along? I thought you were asking about the space of functions $f:\mathbb{R}^n\to\mathbb{R}$ with $\int \lvert f\rvert^p<\infty$ because the term "closed subspace" put my brain in infinite-dimensional mode. So I think I reduced a slightly more complicated question to the one you actually asked before answering. – Noah Stein Feb 9 '12 at 15:24
I am clear now. I hope the fact " All nontrivial $p$-balls are strictly convex, so $(1,1)$ is strictly closer to $x$ than $y$ is unless $y = \hat{y} = (1,1)$." is a consequence of the uniform convexity of the space $L_p(\mathbb{R}^n)$. Thanks, I had a nice time discussing with you. – Ashok Feb 9 '12 at 16:07

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