Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.