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Let $\left( e_n \right)$ be an orthonormal sequence in an inner product space $X$. Then for every $x \in X$, we have $$ \sum_{n=1}^\infty \left\vert \langle x, e_n \rangle \right\vert^2 \ \leq \ \Vert x \Vert^2, $$ where $\Vert \cdot \Vert$ denotes the norm induced by the inner product. This is called the Bessel inequality.

Now how to derive the following inequality, called the Schwarz inequality, from the Bessel inequality?

For any elements $x, y \in X$, $$ \vert \langle x, y \rangle \vert \ \leq \ \Vert x \Vert \cdot \Vert y \Vert. $$

In the finite-dimensional case, this is not so hard as in this case the orthonormal sequence will have only finitely many (distinct) terms, say, $\{e_1, \ldots, e_n\}$, and the last set will then become a basis of $X$.

So every $x, y \in X$ have the unique representations $$x = \sum_{i=1}^n \langle x, e_i \rangle e_i$$ and $$y = \sum_{i=1}^n \langle y, e_i \rangle e_i,$$ respectively, as a linear combination of the $e_i$s.

Then we would have $$\Vert x \Vert^2 = \langle x, x \rangle = \sum_{i=1}^n \left\vert \langle x, e_i \rangle \right\vert^2,$$ $$\Vert y \Vert^2 = \langle y, y \rangle = \sum_{i=1}^n \left\vert \langle y, e_i \rangle \right\vert^2,$$ and $$\langle x, y \rangle = \sum_{i=1}^n \left( \langle x, e_i \rangle \cdot \overline{\langle y, e_i \rangle} \right);$$ so that $$ \begin{align*} \left\vert \langle x, y \rangle \right\vert &= \left\vert \sum_{i=1}^n \left( \langle x, e_i \rangle \cdot \overline{\langle y, e_i \rangle } \right) \right\vert \\ &\leq \sum_{i=1}^n \left( \left\vert \langle x, e_i \rangle \right\vert \cdot \left\vert \langle y, e_i \rangle \right\vert \right) \\ &\leq \sqrt{\left( \sum_{i=1}^n \left\vert \langle x, e_i \rangle \right\vert \right)} \cdot \sqrt{\left( \sum_{i=1}^n \left\vert \langle y, e_i \rangle \right\vert \right) } \\ &= \Vert x \Vert \cdot \Vert y \Vert. \end{align*} $$

Is this reasoning correct?

And if so, then how to tackle the case when $X$ is infinite-dimensional?

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  • $\begingroup$ You are using Schwarz's inequality to prove itself. $\endgroup$
    – Alamos
    Apr 28, 2015 at 6:26
  • $\begingroup$ @Alamos, yes, I'm using the Schwarz equality for real numbers. Do you think even that is not desirable? $\endgroup$ Apr 29, 2015 at 7:31
  • $\begingroup$ No, there would be content in such a proof, but a rewriting of what norm and inner product mean. You can follow any of the many proofs there are that only use the properties of the inner product. $\endgroup$
    – Alamos
    Apr 29, 2015 at 10:55

1 Answer 1

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What you are doing is actually using Schwarz inequality to prove Schawarz inequality.

In the last step you need to use the Holder's Inequality as follows \begin{align} |\langle x,y\rangle|& = \biggl{|}\sum_\limits{n=1}^{\infty}\langle x,e_1\rangle \overline{\langle y,e_i\rangle}\biggr{|}\\ &\leq \sum_\limits{n=1}^{\infty}|\langle x,e_1\rangle|| \overline{\langle y,e_i\rangle}|\\ &\leq \sqrt{\sum_\limits{n=1}^{\infty}|\langle x,e_1\rangle|^2} \sqrt{\sum_\limits{n=1}^{\infty}|\langle y,e_1\rangle|^2} && \text{ (Holder's Inequality)}\\ &\leq \|x\|\|y\| && \text{(Bessel's Inequality)}. \end{align}

$\textbf{EDIT:}$ Sorry for this not helping answer. An easy way out is as follows:

Let $x\in X$ and let $y\neq 0$ be given. Then just take $e=\frac{y}{\|y\|}$. This $\{e\}$ only makes an orthonormal system. So, by Bessel's Inequality it will follow that $|\langle x,e\rangle|^2 \leq \|x\|^2$. Now just multiply both sides by $\|y\|$ to get $|\langle x,y\rangle|^2\leq \|x\|\|y\|$.

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  • $\begingroup$ you'v used something even stronger than what I've used. How do you arrive at the first equality? You see, not every orthonormal sequence is a basis. $\endgroup$ Apr 29, 2015 at 7:37
  • $\begingroup$ @SaaqibMahmuud I have made an edit!!! Sorry for not reading the question completely in first place.......Hope the edit helps!!! $\endgroup$ Apr 29, 2015 at 10:21
  • $\begingroup$ yes, your edit is correct and very easy as well. $\endgroup$ May 27, 2015 at 4:45

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