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If $X$ is a normed space and $Y$ is a finite dimensional subspace, then there exists a continuous linear projection $P$ from $X$ to $Y$. Our teacher gave us the instruction to use the following fact: Let $x_1,\cdots,x_n$ in $X$ be linearly independent. Then there exist $x'_1,\cdots,x'_n$ in $X'$ such that $x'_k(x_r)= \delta_{kr}, 1 \le k,r \le n$. How does one proceed with this assumption? Thank you!

Yuki
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nada
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    Let $x_1,\cdots,x_n$ a basis for $Y$, and define $P:X\to Y$, $P=x_1x'_1+\cdots+x_nx'_n$ – Yuki Jun 10 '12 at 18:55
  • Hello Yuki, please do help me with how one is supposed to use the property to solve the question. Why is it possible to define the mapping P like this? Thank you. – nada Jun 10 '12 at 19:34

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@Yuki gave you the answer.

Define $P:X\to X$ by $Px = \sum x_k'(x) x_k$. Show that $P$ is linear, continuous and $P P = P$.

Potato
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copper.hat
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  • How does one show that P is continuous? And where is the property mentioned in the question used in the construction of P? – nada Jun 10 '12 at 19:52
  • I mean, if x'k(x) equals delta k, so how does that help? – nada Jun 10 '12 at 19:53
  • Also, if we try to show that PP is P, we consider Px = sum x'k(x)xk. This is sum of delta k.xk. Here, both quantities depend on k, so when we want to apply P again on this expression, which term is the new "x"? – nada Jun 10 '12 at 20:09
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    This is homework. You need to do some work. Showing continuity is almost trivial using the triangle inequality. Linearity is trivial. Use the 'orthogonality' property to show $P^2=P$. You have a definition for $P$. Replace the '$x$' by $Px$. Expand the inner sum using a different letter for the inner summation. – copper.hat Jun 10 '12 at 20:29
  • Yes, I showed the linearity and continuity. I am getting stuck at a particular part in the proof of the projection. There are two ways to proceed to show P square is P. The first is the way I did in the prev post. Following your suggestion when I start with P(P(x)) and write this as sum of x'k(Px)xk, then this automatically becomes equal to sum of delta k. xk. Where is the need for the inner summation, then? – nada Jun 10 '12 at 20:40
  • Okay, I think I understand, does the proof go this way: P(Px)) = sum(x'k(Px)xk) = sum(x'k(sum(x'j(x).xj)xk) = sum(xj x'k(sum(x'j(x))xk). Now applying that xj.xk=0 if j not equals k, I don't understand the result. – nada Jun 10 '12 at 20:53
  • $P(Px) = \sum_k x_k^'(Px) x_k = \sum_k x_k^'(\sum_j x_j^'(x) x_j) x_k = \sum_k \sum_j x_j^'(x) x_k^'( x_j) x_k = \sum_k x_k^'(x) x_k = Px$ – copper.hat Jun 10 '12 at 22:12