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In De Boor (1972) it is stated that

Let be $\$ $ a finite dimensional linear space of functions defined on the interval $[a,b]$. We are searching for the best approximation from $\$$ to $g$.

The function $f^*$ is a best approximation from $\$$ to $g$ with respect to the $\mathcal{L}_2$ norm if and only if the function $f^*$ is in $\$$ and the error term $g-f^*$ is orthogonal to $\$$.

To show that the double implication holds the author provides the following statement (I add my own procedure because it's not given in the book).

For any function $f\in \$\ $ we have that $$\|g-f\|_2^2=\|g-f^*+f^*-f\|_2^2=\|g-f^*\|_2^2+2\langle f^*-f,g-f^*\rangle+\|f^*-f\|_2^2$$

If condition is satisfied we have

$$\|g-f\|_2^2=\|g-f^*\|+\|f^*-f\|_2^2\geq\|g-f^*\|_2^2$$ Which proves that $f^*$ is the best least squares approximation in $\$$

If $\langle f,g-f^*\rangle \neq 0$ we have that by letting $tf:=f^*-f$

$$\|g-f\|_2^2=\|g-(f^*+tf)+(f^*+tf)-f\|_2^2$$ $$=\|g-(f^*+tf)\|_2^2+2\langle 2tf,g-f^*-tf\rangle+\|2tf\|_2^2$$

$$=\|g-(f^*+tf)\|_2^2+4\langle tf,g-f^*\rangle$$ Given that for all nonzero $t$ of the same sign as $\langle f,g-f^*\rangle$ and sufficiently close to $0$.

$$2t\langle f,g-f^*\rangle>\|tf\|_2^2$$ This completes the proof since it implies that \begin{equation} \|g-(f^*+tf)\|_2^2<\|g-f^*\|_2^2 \end{equation} and hence $f^*$ is not the best approximation.

I am not sure to understand the reason why it holds that $$2t\langle f,g-f^*\rangle>\|tf\|_2^2$$

Is it because $t^2$ goes to zero faster than $t$. Is there some way to show it more formally? Thanks in advance.

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  • $\begingroup$ I suspect there are some typos, e.g. $\langle f^*-f,g-f^*\rangle$ in the first line after "we have that" and $||g-f||_2^2=||g-f^*||_2^2+||f^*-f||_2^2\geq ||g-f^*||_2^2$ $\endgroup$ – Peter Melech Jan 23 at 16:33
  • $\begingroup$ @PeterMelech I agree with you about the second typo, but the first one,( the inner product) looks okey to me, what do you think? $\endgroup$ – RScrlli Jan 23 at 16:37
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    $\begingroup$ I'm quite sure it has to be $f^*$ instead of $f$ in the right argument of the inner product, similarly: If $\langle f,g-f^*\rangle\neq 0$ some lines under, when showing the other direction $\endgroup$ – Peter Melech Jan 23 at 16:46
  • $\begingroup$ Besides: wonder how" by letting $tf:=f^*-f$" is to be understood $\endgroup$ – Peter Melech Jan 23 at 16:57
  • $\begingroup$ @PeterMelech you are right, I've done a copy and paste from another document and I didn't realize that. Thanks! $\endgroup$ – RScrlli Jan 23 at 16:57
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I would like to suggest a slightly different proof, because I don't feel well with the argument " by letting $tf:=f^*-f$".

Assume $\langle f,g-f^*\rangle\neq 0$ for some $f\in U$ (I call the subspace $U$), then for all $t\in\mathbb{R}$ one has: $$||g-f^*||_2^2=||g-f^*+tf-tf||_2^2=||g-(f^*+tf)||_2^2+2\langle g-(f^*+tf),tf\rangle +||tf||_2^2$$ $$=||g-(f^*+tf)||_2^2+2t\langle g-f^*,f\rangle -2t^2\langle f,f\rangle+||tf||_2 ^2$$ $$=|||g-(f^*+tf)||_2^2+2t\langle g-f^*,f\rangle-t^2||f||_2^2.$$ And now choose $t$ so that $$2t\langle g-f^*,f\rangle-t^2||f||_2^2>0$$ which in case that $t$ has the same sign as $\langle g-f^*,f\rangle$ is equivalent to $$|t|<\frac{2|\langle g-f^*,f\rangle|}{||f||_2^2}.$$ (Note that $\langle g-f^*,f\rangle\neq 0$ is essential to that.) Then: $$|||g-(f^*+tf)||_2^2<||g-f^*||_2^2$$ which contradicts the best-approximation-property.

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  • $\begingroup$ It's much more clear and general as a proof, for sure I will include it in my thesis with a footnote thanking you for the hint $\endgroup$ – RScrlli Jan 24 at 15:27
  • $\begingroup$ Thank You very much! $\endgroup$ – Peter Melech Jan 25 at 7:49
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If $t$ has same sign as $\langle f, g-f^{*}\rangle$ the inequality you want is $2|t||\langle f, g-f^{*}\rangle| >|t|^{2} \|f\|_2^{2}$. This is true whenever $|t| <\frac {2|\langle f, g-f^{*}\rangle|} { \|f\|_2^{2}}$.

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  • $\begingroup$ by taking the absolute value everything makes sense, but the problem is that what I want to show is $2\langle f^*-f,g-f\rangle+\|f^*-f\|_2^2>4\langle tf,g-f^*\rangle$. $\endgroup$ – RScrlli Jan 23 at 10:40
  • $\begingroup$ I've edited the question, showing the first part of the proof. $\endgroup$ – RScrlli Jan 23 at 10:40

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