Possibly a dumb question. In the implicit function theorem we take $F\in C^k(\Lambda\times U, Y)$ with $k\geq 1$, $Y$ is a Banach space, and $\Lambda, U$ are open subsets of Banach spaces $T,X$. If $F(\lambda^*,u^*) = 0$ and $F_u(\lambda^*,u^*)$ is an invertible linear map then there exists neighbourhoods $\Theta$ and $V$ of $\lambda^*$ and $u^*$ and a map $g\in C^k(\Theta,U^*)$ such that $$F(\lambda, g(\lambda))=0$$ for all $\lambda\in\Theta$. What type of convergence does $g(\lambda)\to u^*$ have? For example, if we take $X$ to be $$H_r^1=\{u:\int_0^\infty (u^2+(u')^2)rdr <\infty\}$$ does $g(\lambda)\to u^*$ uniformly? In norm? Just pointwise? By uniformly I mean $|g(\lambda)-u^*|<\epsilon$ for all $r$ when $|\lambda-\lambda^*|<\delta$

  • 1
    $\begingroup$ What do you mean by uniform vs. pointwise? The convergence is in whatever norm the Banach space is supplied with. Maybe an example that illustrates the confusion would be useful? $\endgroup$ – copper.hat May 10 '15 at 20:16
  • $\begingroup$ I'm dealing with a problem right now where $X$ is essentially $H^1$ on the positive real line. So convergence in norm does not imply uniform convergence, which I guess answers the question I wanted to ask. $\endgroup$ – Jnkiejim May 10 '15 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.