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I'm trying to prove that $\sum_{j=1}^n\langle x_j,y_j\rangle \leq \sqrt{ \sum_{j=1}^n\langle x_j,x_j\rangle} \sqrt{\sum_{j=1}^n \langle y_j,y_j\rangle}$, where $\left(V, \langle \cdot , \cdot \rangle \right)$ is an inner product space and $x_1, x_2, ...$ and $y_1, y_2,...$ are sequences of elements of $V$.

My attempt: $$\sum_{j=1}^n\langle x_j,y_j\rangle \leq \sum_{j=1}^n\langle x_j,x_j\rangle^{1/2}\langle y_j,y_j\rangle^{1/2} \leq \left(\sum_{j=1}^n \langle x_j,x_j\rangle\right)^{1/2} \left(\sum_{j=1}^n \langle y_j,y_j\rangle\right)^{1/2}.$$

I'm not sure if the first inequality holds for an arbitrary inner product space. The second inequality is based on Cauchy. Am I correct?

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    $\begingroup$ "I'm not sure if the first inequality holds for an arbitrary inner product space" Yes, it's exactly the Cauchy inequality. $\endgroup$ May 19, 2018 at 19:17
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    $\begingroup$ Actually you are applying Cauchy twice, for different spaces and inner products. But it's still OK. $\endgroup$ May 19, 2018 at 19:21
  • $\begingroup$ @Logic_Problem_42 Thank you. I applied Cauchy twice for different inner product spaces so I think I also need to show $[a,b] = \sum_{j=1}^n \langle a,a \rangle^{1/2} \langle b,b \rangle^{1/2}$ is also an inner product? $\endgroup$
    – user547265
    May 19, 2018 at 19:33
  • $\begingroup$ Actually You don't. You use in this step the standard inner product from $R^n$: $\langle x, y\rangle=\sum_{k=1}^n x_iy_i$. And the Cauchy inequality is then $|\sum_{k=1}^n x_iy_i| \le (\sum_{k=1}^n x_i^2)^{1/2}(\sum_{k=1}^n y_i^2)^{1/2}$. You simply use it for vectors $(\langle x_1, x_1\rangle^{1/2},...,\langle x_n, x_n\rangle^{1/2})$ and $(\langle y_1, y_1\rangle^{1/2}),...,\langle y_n, y_n \rangle^{1/2})$ instead of $x$ and $y$. $\endgroup$ May 19, 2018 at 19:38
  • $\begingroup$ @Logic_Problem_42 The problem doesn't say that $\langle \cdot, \cdot \rangle$ is from $V\times V \to \Bbb R^n$ so I think the space of the domain can be any field. Is it correct to say that we use the standard inner product from $\Bbb R^n$? $\endgroup$
    – user547265
    May 19, 2018 at 20:41

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I probably write it as the answer, I think it would be clearer this way.

So, you have some scalar product $\langle,\rangle$ on $V\times V$. As with every scalar product, you have the Cauchy inequality: $$|\langle x,y\rangle| \le \langle x,x\rangle^{1/2}\langle y,y\rangle^{1/2}.$$

It is valid for all possible $x,y$, so also for $x_i,y_i$. Hence you have for each $i$: $$|\langle x_i,y_i\rangle| \le \langle x_i,x_i\rangle^{1/2}\langle y_i,y_i\rangle^{1/2}.$$ Summing up you then have $\sum_{i=1}^n|\langle x_i,y_i\rangle| \le \sum_{i=1}^n\langle x_i,x_i\rangle^{1/2}\langle y_i,y_i\rangle^{1/2}$.

In the next step you consider another space: $\Bbb R^n$. In this space there is the standard scalar product and then the Cauchy inequality with respect to this product. It says that for each $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ you have $$\sum_{k=1}^n |x_iy_i|\le \left(\sum_{i=1}^n |x_i|^2\right)^{1/2} \left(\sum_{i=1}^n |y_i|^2\right)^{1/2}.$$

As it holds for all $x,y$, it also holds for $x=(\langle x_1,x_1\rangle^{1/2},...,\langle x_n,x_n\rangle^{1/2})$ and $y=(\langle y_1,y_1\rangle^{1/2},...,\langle y_n,y_n\rangle^{1/2})$ which are simply the vectors from $\Bbb R^n$. Now for these concrete vectors, Cauchy says that $$\sum_{i=1}^n \langle x_i,x_i\rangle^{1/2} \langle y_i,y_i\rangle^{1/2}\le \left(\sum_{i=1}^n \langle x_i,x_i\rangle\right)^{1/2} \left(\sum_{i=1}^n \langle y_i,y_i\rangle\right)^{1/2}.$$ This proves the second inequality.

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