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Why is the LHS of this inequality a series uniform and absolute convergence?

It's about integral operators... let's see $f\in L^2(\mu)$, $\lambda_n$ a sequence of eigenvalues, $\varphi_n$ a sequence of eigenvectors, X a compact space and $M$ just a constant. I don't want to prove this inequality just need to deduct the convergence of first series. –

$$\sum_{n=p}^q\left|\lambda_n\left(\int_{X}f(y)\;\overline{\varphi_n}(y)\;d\mu(y) \right)\varphi_n(x)\right|\leq M\cdot \left({\sum_{n=p}^q\left|\int_{X}f(y)\;\overline{\varphi_n}(y)\;d\mu(y)\right|^2}\right)^{1/2}$$

It's useful to say I guess that $$\langle f,\varphi_n\rangle=\int_{X}f(y)\,\overline{\varphi_n}(y)\;d\mu(y)$$

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    $\begingroup$ Please edit your question to include your own thoughts and efforts, where you encountered this problem, and what tools you have available. In particular, it would be helpful to actually define the functions involved, and ask a question that's a lot more clear than this one. $\endgroup$ – user296602 Apr 20 '18 at 22:20
  • $\begingroup$ It's about integral operators... let's see $f\in L^2(\mu)$, $\lambda_n$ a sequence of eigenvalues and $\varphi_n$ a sequence of eigenvectors. I don't want to prove this inequality just need to assume the convergence of first series. $\endgroup$ – Andrew Apr 20 '18 at 22:25
  • $\begingroup$ Please use MathJax to transcribe the graphics $\ddot\smile$ $\endgroup$ – gen-z ready to perish Apr 20 '18 at 22:37
  • $\begingroup$ I have rewritten everything. $\endgroup$ – Andrew Apr 20 '18 at 23:11
  • $\begingroup$ As, you see, some people here are trigger-happy to close questions by newcomers, but then they don't come back to re-open it when edited. Anyway, the answer to your question is that the series on the right is convergent (the full series converges to $\|f\|_2$, by Parseval's equality). If you have the inequality of all $p,q$ then you are showing that the tails of the series on the left go to zero, which means that the series converges. $\endgroup$ – Martin Argerami Apr 21 '18 at 15:33
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A series $\sum_n a_n$ converges (be definition) if the sequence of partial sums $$ \left\{\sum_{n=1}^ma_n\right\}_m $$ converges. This sequence of partial sums converges if and only if it is Cauchy: that is, given $\varepsilon>0$ there exists $m_0$ such that whenever $q>p\geq m_0$, $$ \left|\sum_{n=1}^p a_n-\sum_{n=1}^q a_n\right|<\varepsilon. $$ Now note that $$ \sum_{n=1}^p a_n-\sum_{n=1}^q a_n=\sum_{n=p+1}^q a_n. $$ The inequality you have shows that the tails of the series on the left are bounded by the tails of the series on the right. The series on the right is convergent (Parseval).

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