Inequality involving series using Cauchy-Schwartz

[powerof]

Homework Statement

[/B]
Prove the following:

[itex]\sum_{k=1}^{\infty}a_{k}^{2} \leq \left ( \sum_{k=1}^{\infty}a_{k}^{2/3} \right )^{1/2} \left ( \sum_{k=1}^{\infty}a_{k}^{4/3} \right )^{1/2}[/itex]

Homework Equations

[/B]
The following generalization of Cauchy-Schwarz present in the text (containing the problem) is useful (assume all the sequences are so that the series converge):

[itex]\sum_{k=1}^{\infty}a_{k}b_{k} \leq \left ( \sum_{k=1}^{\infty}a_{k}^{2} \right )^{1/2} \left ( \sum_{k=1}^{\infty}b_{k}^{2} \right )^{1/2}[/itex]

The Attempt at a Solution

[/B]
I have tried thinking of defining ##a_{k}## and ##b_{k}## as powers of an identical sequence: ##a_{k} = \phi_{k}^{n}## and ##b_{k}=\phi_{k}^{m}##. Substituting into Cauchy-Schwartz I may get an idea of what n and m should be:

[itex]\sum_{k=1}^{\infty}\phi_{k}^{n}\phi_{k}^{m} \leq \left ( \sum_{k=1}^{\infty}(\phi_{k}^{n})^{2} \right )^{1/2} \left ( \sum_{k=1}^{\infty}(\phi_{k}^{m})^{2} \right )^{1/2}\Rightarrow 2n=2/3,2m=4/3 \Rightarrow n=\frac{1}{3},m=\frac{2}{3}[/itex]

With these values of n and m I am left with:

[itex]\sum_{k=1}^{\infty}\phi_{k} \leq \left ( \sum_{k=1}^{\infty}\phi_{k}^{2/3} \right )^{1/2} \left ( \sum_{k=1}^{\infty}\phi_{k}^{4/3} \right )^{1/2}[/itex]

I don't get the ##\phi_k ^2## on the left side but at least I have the right powers on the right side.

Now if my method is correct I need to show that ##\sum_{k=1}^{\infty}a_{k}^{2}## is bounded by ##\sum_{k=1}^{\infty}a_{k}## (note that I have substituted phi with a but it's just names so it doesn't matter) because the last is bounded by the right side thus proving the initial statement. In the infinite it's somewhat intuitive since the terms get closer and closer to 0 (necessary for convergence) and a small number, x, when squared gets even lower in value so x>x^2 when x small. But this trend is assured in the limit, in the infinite. It says nothing about what the first term, the second, and so on is so I cannot use that reasoning here.

I am stuck so I would appreciate some hints as to what I can do or different approaches to this problem.

Thank you for your time and have a nice day.

 

[pasmith]

I think you have the intended method, but I can't see how to get the right hand side correct if you fix [itex]n + m = 2[/itex] as the left hand side requires.

Certainly you cannot conclude that [itex]\sum_{k=1}^\infty \phi_k^2 \leq \sum_{k=1}^\infty \phi_k[/itex]: take [itex]\phi_k = -1/k^2[/itex] for example.

 

[mfb]

I guess this is a typo in the problem statement and the left side should be a_k. You cannot prove something wrong.
Consider a₁=8, a_k=0 for all other k. The left side is 64, the right side is 8.

 

[powerof]

Thank your for your help. The counterexamples definitely make this inequality impossible, so a typo it must be. Have a nice day!

 

[Svein]

Cauchy-Schwarz for ℝ^N states [itex]\vert \sum_{j=1}^{N}u_{j}v_{j}\vert \leq \sqrt{\sum_{j=1}^{N}u_{j}^{2}}\cdot \sqrt{\sum_{j=1}^{N}v_{j}^{2}} [/itex]. Now, let [itex] u_{k}=a_{k}^{\frac{1}{3}}[/itex] and [itex] v_{k}=a_{k}^{\frac{2}{3}}[/itex]. Then [itex]u_{k}v_{k}=a_{k}^{\frac{1+2}{3}}=a_{k} [/itex]. Cauchy-Schwarz - still for ℝ^N - gives [itex]\vert \sum_{k=1}^{N}a_{k}\vert \leq \sqrt{\sum_{k=1}^{N}a_{k}^{\frac{2}{3}}}\cdot \sqrt{\sum_{k=1}^{N}a_{k}^{\frac{4}{3}}} [/itex].

1. The answer is different from the problem statement
2. There is no guarantee that the sums under the square root converge as N→∞