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help me. someone who can help me? spaces is inner product. It is section 3.2, issue 4 introduction to functional analysis book author Kreyszig

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  • $\begingroup$ Perhaps use linearity properties of the inner product, and the general Cauchy-Schwartz? You can try setting the problem up and giving your work for more help, or that just might be enough for you to solve it on your own. $\endgroup$ – Michael May 26 '15 at 22:32
  • $\begingroup$ Or just use the continuity of the inner product. $\endgroup$ – zuggg May 26 '15 at 22:36
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For arbitrarily $\epsilon > 0$ there exists an $N$ such that $n > N$ implies $\left|\langle y, x\rangle\right| = \left|\langle y, x_n - x \rangle\right| \leq \| y \| \| x_n - x \| < \epsilon$ . Hence ...

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  • $\begingroup$ because equality between the first two expressions given? $\endgroup$ – daniel May 26 '15 at 22:41
  • $\begingroup$ Because $\langle y, x_n \rangle = 0$ by hypothesis and hence $\left|\langle y, x_n - x\rangle \right| = \left|\langle y, x_n \rangle - \langle y, x \rangle\right| = \left|\langle y, x \rangle\right| $ $\endgroup$ – Simon S May 26 '15 at 22:42
  • $\begingroup$ @SimonS : You actually manually put a space between $\|y\|$ and $\|x_n-x\|$. If you code it as \|y\|, then not only do you see $\|y\|$, with the vertical bars closer together than if you code it as ||y||, but also that extra space appears there automatically; it's built in to the software. I changed it. $\endgroup$ – Michael Hardy May 27 '15 at 3:44
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Which kind of convergence are we talking about. If $y=(y_1,\ldots,y_d)$ then it doesn't matter, but if $y=(y_1,y_2,y_3,\ldots)$ with infinitely many components, then it does.

What does $x_n \to x$ mean? One characterization is $\|x_n-x\|\to 0$, i.e. convergence of a sequence of vectors is defined by convergence of a sequence of numbers. And this is equivalent to $\|x_n-x\|^2\to 0$, and that is $(x_n-x)\cdot(x_n-x)$, the dot-product, approaches $0$ as $n\to\infty$.

So we have $y\cdot x_n=0$ for every $n$. We want to know whether $y\cdot x=0$.

$$ y\cdot x = y\cdot((x-x_n)+x_n) = y\cdot(x-x_n) + y\cdot x_n = y\cdot(x-x_n). $$ How can $y\cdot(x-x_n)$ remain equal to $y\cdot x$ as $n$ changes, when $y\cdot x$ does not depend on $n$? That can happen only if $y\cdot(x-x_n)$ does not depend on $n$, i.e. $y\cdot x = y\cdot x_n$ for every $n$. But $y\cdot x_n$ is $0$. So $y\cdot x$ is $0$.

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  • $\begingroup$ thank you very much indeed . $\endgroup$ – daniel May 26 '15 at 22:51

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