I urgently need help on the following problem:

Problem: Show that if $\displaystyle\sum_{n=1}^{\infty }f_{n}^{2}\left ( x \right )$ is pointwise convergent on $A$, and if $\sup_{x\in A} \left ( \sum_{n=1}^{\infty }f_{n}^{2}\left ( x \right ) \right )< \infty $ and $\sum_{n=1}^{\infty }c_{n}^{2}$ converges, then $\sum_{n=1}^{\infty } c_{n}f_{n}\left ( x \right )$ converges uniformly on $A$.

Here is what I did : My purpose is to prove uniform convergence using the Cauchy criterion for uniform convergence. Using the Cauchy inequality, we get: $$\sup_{x\in A}\left | \sum_{k=n}^{n+m } c_{k}f_{k}\left ( x \right ) \right |\leqslant \left ( \sum_{k=n}^{n+m }c_{k}^{2}\right )^{\frac{1}{2}} \sup_{x\in A}\left ( \sum_{k=n}^{n+m }f_{k}^{2}\left ( x \right ) \right )^{\frac{1}{2}}.$$

Now, let $\epsilon > 0$ be arbitrary. Since $\left (\sum_{k=1}^{\infty }c_{k}^{2}\right)$ is convergent, then by Cauchy condition, there exists $n_{0}$ in $N$ such that for every $m> n_{0}$, we have : $\left ( \sum_{k=n}^{n+m }c_{k}^{2}\right )< \frac{\epsilon ^{2}}{M^{2}}$. Also, let $\sup_{x\in A} \left (\sum_{n=1}^{\infty }f_{n}^{2}\left ( x \right) \right )=M^{2}$ then we get that there is $n_{0}$ in $N$ such that for every $m> n_{0}$, we have: $$\sup_{x\in A}\left | \sum_{k=1}^{n+m } c_{k}f_{k}\left ( x \right ) \right |< \frac{\epsilon }{M} M=\epsilon $$ which proves uniform convergence according to Cauchy Criterion for uniform convergence.

Please let me know if my proof is correct? If not let me know the correct proof and/or which details I should add/remove in order to finish off my proof? Thanks a lot.

  • 2
    $\begingroup$ I hope, you meant to say $sup_{x\in A}\left | \sum_{k=n}^{n+m } c_{k}f_{k}\left ( x \right ) \right |$ wherever you have $sup_{x\in A}\left | \sum_{k=1}^{n+m } c_{k}f_{k}\left ( x \right ) \right |$. Then I don't see any problem with the solution. $\endgroup$ – Ashok Nov 17 '11 at 7:16
  • $\begingroup$ Yes, That's what I meant.Now, I corrected the original statement. $\endgroup$ – M.Krov Nov 17 '11 at 14:11
  • $\begingroup$ This sounds like a job for Superman - I mean the Weierstrass M-test $\endgroup$ – kahen Nov 30 '11 at 13:39

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