Let $d, n \in \mathbb{N}$. Moreover, let $D \subset \mathbb{R}^d$ be compact and denote with $\mathcal{C}(D, \mathbb{R}^n) $ the set of continuous functions from $D$ to $\mathbb{R}^n$. Then $\mathcal{C}(D, \mathbb{R}^n) $ is a Banach space with the usual maximum norm $\lVert f\rVert_{\infty} := \max_{x \in D} \lVert f(x) \rVert $.

Now choose $K>0$ and $l \in \mathbb{N}$. The map $$ j : [-K,K]^l \rightarrow \mathcal{C}(D, \mathbb{R}^n) $$ $$ j(\theta) = \Phi_{\theta} $$ parametrizes a family of continuous functions $\Phi_{\theta}$ in such a way that, for each fixed $x \in D$, the map $$h_x : [-K,K]^l \rightarrow \mathbb{R}^n $$ $$h_x(\theta) = \Phi_{\theta}(x) $$ is continuous in $[-K,K]^l$ and continuously differentiable in $(-K,K)^l$.

From this, I want to conclude that $j$ is continuous. Is this true? If not, what additional assumptions would make this true?

I have tried using the sequence criterion of continuity for $j$, i.e. $\theta_n \rightarrow \theta$ should imply that $j(\theta_n) \rightarrow j(\theta)$. I then tried to show $j(\theta_n) \rightarrow j(\theta)$ by making use of the multidimensional mean-value theorem, but so far I have not been successful.

I am grateful for any advice!

Kind regards and thank you very much,


  • $\begingroup$ Your assumption means that $j$ is continuous if the range $C(D,\mathbb R^n)$ is endowed with the topology of pointwise convergence (which is much coarser that the norm topology). If the range of $j$ were (relatively) norm compact the topologies would coincide. You could try to apply Arzela-Ascoli to get the norm-compactness (i.e, you need equicontinuity of the family $\{\Phi_\theta\}$). Conversly, if $j$ is norm continuous then $\{\Phi_\theta\}$ is compact as the continuous image of the compact space $[-K,K]^l$. $\endgroup$ – Jochen Feb 14 at 13:36

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.