Is this mapping from a hypercube into a Banach space of continuous functions itself continuous?

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,

Joker

• 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 $[-K,K]^l$. – Jochen Feb 14 at 13:36