I want to show that for any continuous function $f:R^N\to R^V$ of at most linear growth, i.e. $$\exists C<\infty\forall x\in R^N\ ||f(x)||\leq C(1+||x||), $$ there exists a sequence $f_n:R^N\to R^V,n\geq1,$ such that
- $f_n$'s are of at most linear growth, uniformly in $n$.
- $f_n$'s are Lipschitz continuous.
- $f_n\to f$ locally uniformly on $R^N$.
As a hint suggested I defined convolution functions $g_n(x)=n^N\int_{R^N}f(y)\zeta(n(x-y))dy,$ for some $\zeta\in C^\infty(R^N,R)$ with $\zeta\geq0,$ supp$\zeta\subset$ unit ball, and $\int_{R^N}\zeta(x)dx=1$ and I showed that $g_n$'s satisfy points 1. and 3. but only locally Lipschitz continuity. At this point I should modify the $g_n$'s to get globally Lipschitz condition, but I don't know how to proceed. I think I have to choose nice compact $K\subset R^N$ (large enough to not lose the convergence) and set $f_n(x)=g_n(x)1_{K}+something\cdot1_{K^c}$, (similarly as this question Extending a $k$-lipschitz function), but I'm not sure. Thanks in advance.