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Let $\Omega$ be Hausdorff locally compact and let $X$ be a Banach Space with norm $\| \cdot \|$. Denote by $C_c (\Omega, X)$ the space of continuous function $f: \Omega \to X$ with compact support $K_f$.

I'm asked to proof that:

  1. Given $f \in C_c(\Omega, X)$ and $\epsilon > 0$, exists $U_1,...,U_n \subset \Omega$ open sets such that $K_f \subset \bigcup_{i=1}^n U_i$ and $\forall x,y \in U_i, \, \|f(x) - f(y) \| < \epsilon, \, \forall i = 1,...,n$.

  2. Given $f \in C_c(\Omega, X)$ and $\epsilon > 0$, consider $U_1,...,U_n \subset \Omega$ as in 1. Then, for all $j=1,..,n$, exists $f_j \in C_c(U_j, \mathbb C)$ and $x_j \in X$ such that $$ \left \| f - \sum_{j=1}^n f_j(\omega) x_j \right \| < \epsilon. $$

The first question, I managed to proof. I'm stucked in the second one. I got the hint to use the unity partition of $K_f$, i.e for each $j = 1,...,n$, take $g_j: U_j \to \mathbb C$ with compact support such that $g_1(\omega) + ... + g_n(\omega) = 1, \forall \omega \in K_f$.

I've tried to put $f_j = g_j$ and $x_j = f(\omega_j)$ for some $\omega_j \in U_j$, however it didn't work.

Help?

Community
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user 242964
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1 Answers1

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Your Ansatz to have $f_i$ be a partition of unity of $K_f$ and $x_i=f(y_i)$ for some $y_i\in U_i$ is correct.

If $x\in U_i$ then $\|f(y)-f(y_i)\|<\epsilon$ from part $1$. Since if $y\notin U_i$ you have $f_i(y)=0$ you get for any $y$: $$\sum_{i=1}^n f_i(y)\,\|f(y)-f(y_i)\|≤\sum_{i=1}^n f_i(y)\,\max_{i: y\in U_i}(\|f(y)-f(y_i)\|)<\sum_{i=1}^n f_i(y)\,\epsilon= \epsilon$$ Thus $$\left\|f(y)-\sum_{i=1}^n f_i(y)f(y_i)\right\|≤\sum_{i=1}^nf_i(y)\|f(y)-f(y_i)\|<\epsilon$$ Taking the supremum over all $y$ gives the result.

s.harp
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