Approximation of bounded and continuous mappings Does anyone know if we can approximate a bounded (i.e. bounded sets in V are mapped to bounded sets in V': for every bounded $U\subseteq V$ and $x\in U$, there exists $K_U>0$ such that $\|f(x)\|_{V'}\leq K_U$, or, in particular, for an $r>0$ and $\|x\|_V\leq r$, $\|f(x)\|_{V'}\leq K_r$) and continuous mapping $f\colon V\rightarrow V'$, where $V,V'$ are real Hilbert spaces, by a sequence of Lipschitz mappings such that the sequence converges uniformly to $f$ on compact sets?
Thank you very much for any help.
Possible weak answer:
The following is my attempt to answer the question. Well, I only got Lipschitz-on-bounded-and-closed-sets sequence instead of globally Lipschitz. It will be appreciated if someone can check my argument, and if one can also point me to the globally Lipschitz sequence.
So, we will construct a sequence $(f_n)_{n\in\mathbb{N}}$ such that each $f_n$ is Lipschitz on bounded and closed sets, and such that $\underset{n\rightarrow\infty}{\lim} f_n=f$ uniformly on compact sets. For any $n$, take the usual bump function $\rho_n\colon\mathbb{R}^n\rightarrow\mathbb{R}$ such that 
support$(\rho_n)=${$\xi\in\mathbb{R}^n\colon\|\xi\|\leq\frac{1}{n}$}, and $\int_{\mathbb{R}^n}\rho_n(\xi)d\xi=1$.
Take any orthonormal basis of $V$, $(e_k)_{k\in\mathbb{N}}$, and define the operator $Q_n\colon V\rightarrow \mathbb{R}^n$ through
$Q_n x:=(<x,e_1>,\ldots,<x,e_n>)$
for every $x\in V$. Then, define $f_n$ through
$f_n(x):=\int_{\mathbb{R}^n}\rho_n(\xi-Q_n x)f(\sum_{k=1}^n\xi_k e_k)d\xi$.
We will first show that $f_n$ is Lipschitz. Denote by $B(x_0,r)$ the closed ball in $\mathbb{R}^n$ with center $x_0$ and radius $r$. Also for brevity, denote $B^n_{x_0}$ as a closed ball of center $x_0$ with radius $\frac{1}{n}$ in $\mathbb{R}^n$. Well, since a bump function is smooth and has compact support, it is globally Lipschitz (with Lipschitz constant $\text{Lip}(\rho_n)$) (thank you for Igor Rivin and copper.hat for pointing this out in another post), and, therefore, for any $\|x\|_V,\|y\|_V\leq r$ we have
\begin{eqnarray}
\|f_n(x)-f_n(y)\|_{V'}&&=\|\int_{B^n_x\cup B^n_y}(\rho_n(\xi-Q_n x)-\rho_n(\xi-Q_n y))f(\sum_{k=1}^n\xi_k e_k)d\xi\|_{V'}\\
&&\leq \text{Lip}(\rho_n)\|Q_n(x-y)\|_{\mathbb{R}^n}\,.\,K_{r+\frac{1}{n}}
\,.\,2\text{Volume}(B(0,\frac{1}{n}))\\
&&\leq \text{Lip}(\rho_n)\,.\,K_{r+\frac{1}{n}}
\,.\,2\text{Volume}(B(0,\frac{1}{n}))\,.\,\|x-y\|_V
\end{eqnarray}
where the last inequality follows from Parseval's and where $K_{r+\frac{1}{n}}$ is the bound (see description of boundedness above) that comes from the boundedness of $f$.
Now we will show the uniform convergence. We have for every compact set $U$ and any $x\in U$,
\begin{eqnarray}
\|f(x)-f_n(x)\|_{V'}&&=\|\int_{B^n_x}\rho_n(\xi-Q_n x)(f(x)-f(\sum_{k=1}^n\xi_k e_k))d\xi\|_{V'}\\
&&=\|\int_{B^n_0}\rho_n(\xi)(f(x)-f(\sum_{k=1}^n(<x,e_k>+\xi_k) e_k))d\xi\|_{V'}\\
&&\leq \int_{B^n_0}\rho_n(\xi)\|f(x)-f(\sum_{k=1}^n(<x,e_k>+\xi_k) e_k)\|_{V'}d\xi,
\end{eqnarray}
Using the compactness of the set (please verify if the following set is indeed compact)
$G:=U\cup(\bigcup_{n=1}^\infty\left\lbrace \sum_{k=1}^n(<x,e_k>+\xi_k) e_k\colon x\in U\text{ and }\sum_{k=1}^n\xi_k^2\leq\frac{1}{n^2}\right\rbrace)$
and the fact that $f$ is uniformly continuous over $G$, we can therefore imply that
\begin{eqnarray}
\underset{x\in U}{\sup}\|f(x)-f_n(x)\|_{V'}&&\leq \underset{\substack{\|x-y\|_V\leq\delta_n\\x,y\in G}}{\sup}\|f(x)-f(y)\|_{V'}\rightarrow 0
\end{eqnarray}
where $\delta_n$ decreases to $0$, and indeed we can take $\delta_n$ as (please verify)
\begin{eqnarray}
\delta_n:=\underset{x\in U}{\sup}\|x-\sum_{k=1}^n<x,e_k>e_k\|_V+\frac{1}{n},
\end{eqnarray}
which is indeed finite for every $n$, as $U$ is compact and therefore bounded.

On the compactness of $G$:
We only need to show the compactness of 
$G':=\cup_{n\in\mathbb{N}}G'_n$,  $G'_n:=\left\lbrace \sum_{k=1}^n(<x,e_k>+\xi_k) e_k\colon x\in U,\quad\xi:=(\xi_1,\ldots,\xi_n)\in B^n_0\right\rbrace$
First, note that for any $n$, $G'_n$ is compact as both $U$ and $B^n_0$ are compact. Now take any sequence $(u_n)_{n\in\mathbb{N}}\subseteq G'$. Then, any $u_n$ is of the form
$u_n=\sum_{k=1}^{l_n}(<x_n,e_k>+\xi^{(n)}_k) e_k$, $x_n\in U$, $\sum_{k=1}^{l_n}(\xi^{(n)}_k)^2\leq\frac{1}{{l_n}^2}$
First of all, if $\sup_n l_n<\infty$, then $(u_n)\subseteq \cup_{n=1}^{\sup_n l_n}G'_n$, and therefore is compact since $\cup_{n=1}^{\sup_n l_n}G'_n$ is a finite union of compact sets.
So, assume that $\lim\sup_n l_n=\infty$. Since $(x_n)_{n\in\mathbb{N}}\subseteq U$, select a subsequence that converges. Assume we have relabeled $(u_n)_{n\in\mathbb{N}}\subseteq G'$ to be the corresponding subsequence. Then, we only need to show that it is Cauchy. For every $m,n\in\mathbb{N}$, we have (suppose wlog $l_m<l_n$) by Minkowski's inequality,
\begin{eqnarray}
\|u_n-u_m\|_{V}&&=\sqrt{\sum_{k=1}^{l_m}(<x_n-x_m,e_k>+(\xi^{(n)}_k-\xi^{(m)}_k))^2+\sum_{k=l_m+1}^{l_n}(<x_n,e_k>+\xi^{(n)}_k)^2}\\
&&\leq\|x_n-x_m\|_V+\sqrt{\sum_{k=l_m+1}^{l_n}<x_n,e_k>^2}+\frac{1}{l_n}+\frac{1}{l_m}\\
\end{eqnarray}
and the rest is obvious by taking $\lim\sup$ on both sides as $m,n\rightarrow\infty$.
 A: It seems the following. 
Proposition. Let $V$ be a separable real Hilbert space, $V’$ be a normed space and $f:V\to V’$ be a continuous map. Then there exists a sequence $\{f_n\}$ of bounded Lipschitz maps from $V$ to $V’$ converging uniformly to $f$ on compact sets.
Proof. It suffices to consider the case $V=\ell_2$. Let $n$ be a natural number. We shall consider the space $\mathbb R^n$ as naturally embedded into $\ell_2$ and put $\mathbb Z/n=\{x\in\mathbb R: nx\in\mathbb Z\}$. At first we define a map $g_n^*:(\mathbb Z/n)^n\to V’$ by putting $g_n^*(x)=f(x)$ if $\|f(x)\|\le n$ and $g_n^*(x)=nf(x)/\|f(x)\|$, if $\|f(x)\|\ge n$. Next we affinely extend the map $g_n^*$ to a map $f_n^*:\mathbb R^n\to V’$ as follows. At first we define a map $\lfloor\cdot\rfloor_n:\mathbb R\to \mathbb Z/n$ by putting $\lfloor x\rfloor_n=\lfloor nx\rfloor/n$ for each $x\in\mathbb R$. Let $x=(x_i)\in\mathbb R^n$. Then the point $x$ is contained in a cube $C_x=\prod_{i=1}^n [\lfloor x_i \rfloor_n; \lfloor x_i \rfloor_n+1/n]$. Put $f_n^*(x)=\sum\{\lambda(d,x)g_n^*(d): d=(d_i)$ is a vertex of the cube $C_x\}$, where $\lambda(d,x)=\prod_{i=1}^n \mu_i(d_i,x_i)$, where $\mu_i(d_i,x_i)=n(x_i-\lfloor x_i \rfloor_n)$  if $d_i=\lfloor x_i \rfloor_n+1/n$ and $\mu_i(d_i,x_i)=1-n(x_i-\lfloor x_i \rfloor_n)$  if $d_i=\lfloor x_i \rfloor_n$. We remark that $f_n^*(x)\in\operatorname{conv}\{g_n^*(d): d$ is a vertex of the cube $C_x\}$ by the construction. We claim that the map $f_n^*$ is Lipschitz. Indeed, let $x,y\in\mathbb R^n$. Since the segment $[x,y]$ is contained in a finite union of the cubes $C_z$ where $z\in \mathbb R^n$, it suffices to consider the case when $C_x=C_y$. In this case $\lfloor x_i \rfloor_n=\lfloor y_i \rfloor_n$ for eac $1\le i\le n$. Then 
$$\left\| f_n^*(x)- f_n^*(y)\right\|=\left\| \sum\{\lambda(d,x)g_n^*(d)- \lambda(d,y)g_n^*(d): d=(d_i)\mbox{ is a vertex of the cube }C_x\}\right\|\le \sum_d |\lambda(d,x)–\lambda(d,y)|\|g_n^*(d)\| \le \sum_d\left|\prod_{i=1}^n \mu_i(d_i,x_i)– \prod_{i=1}^n \mu_i(d_i,y_i)\right|n\le^* \sum_d\sum_i n|x_i-y_i|n\le 2^nn^3\|x-y\|.$$ 
To prove the inequality marked (*) we remark that $0\le\mu_i(d_i,x_i), \mu_i(d_i,y_i)\le 1$ and $|\mu_i(d_i,x_i)- \mu_i(d_i,y_i)|=n|x_i-y_i|$ for every $i$. But if  $0\le a_i, b_i\le 1$ then $|a_1a_2\dots a_n- b_1b_2\dots b_n|=|a_1a_2\dots a_{n-1}a_n- a_1a_2\dots a_{n-1}b_n+
a_1a_2\dots a_{n-1}b_n- a_1a_2\dots b_{n-1}b_n+\dots+ a_1b_2\dots b_{n-1}b_n- b_1b_2\dots b_{n-1}b_n |\le
|a_1a_2\dots a_{n-1}a_n- a_1a_2\dots a_{n-1}b_n|+|a_1a_2\dots a_{n-1}b_n- a_1a_2\dots b_{n-1}b_n|+\dots+|a_1b_2\dots b_{n-1}b_n- b_1b_2\dots b_{n-1}b_n|\le |a_n-b_n|+|a_{n-1}- b_{n-1}|+\dots+|a_1- b_1|$. 
At last, put $f_n=f^*_np_n$, where $p_n:\ell_2\to\mathbb R^n$ is the orthogonal projection. Since the map $p_n$ is Lipschitz, the map $f_n$ is Lipschitz too.
We are going to prove that a sequence $\{f_n\}$ converges uniformly to $f$ on compact sets.
Let $K\subset\ell_2$ be a compact set and $\varepsilon>0$ be an arbitrary real. Each point $x\in K$ has an open neighborhood $Ox$ such that $\|f(x)-f(y)\|<\varepsilon$ for each point $y\in Ox$. We remark that the function $f$ is bounded on the set $\bigcup\{Ox:x\in K\}$. Therefore there exists a number $N_1$ such that $\|f(y)\|\le n$ for each number $n>N_1$ and each point $y\in\bigcup\{Ox:x\in K\}$. 
We claim that there exists a number $\delta>0$ such that for each point $y\in K$ there exist a point $x\in K$ such that $z\in Ox$ provided $z\in\ell_2$ and $\|z-y\|<\delta$. Indeed, for every $x\in K$ take an $\delta_x>0$ such that the open ball $B(x,2\delta_x)$ is contained in $Ox$. The open cover $\{B(x,\delta_x):x\in K\}$ has a finite subcover, that is, there exists a finite set $F\subset K$ such that $K\subset\bigcup \{B(x,\delta_x):x\in F\}$. One can readily see that the number $\delta=\min\{\delta_x:x\in F \}$ has the required property.
For each point $x\in K$ put $h_n(x)=\|x-p_n(x)\|$. Since $x\in\ell_2$, we see that $p_{n+1}(x)\le p_n(x)$ for each $n$ and the sequence $h_n(x)$ converges pointwisely to zero for each point $x\in K$. Then Dini Theorem (see, for instance, [L. 3.2.18, Eng]) implies that the sequence $h_n(x)$ converges uniformly to zero. Therefore there exists a number $N_2$ such that $\|x-p_n(x)\|<\delta/2$ for each number $n>N_2$ and each point $x\in K$. 
Choose a number $N_3\ge \max\{N_1,N_2\}$ such that $1/\sqrt{N_3}<\delta/2$. Now let $x\in K$ be an arbitrary point and $n>N_3$ be an arbitrary number. Since $\|x-p_n(x)\|<\delta/2$ and $1/\sqrt{N_3}<\delta/2$ then the cube $C_{p_n(x)}\subset\mathbb R^n$ is contained in the ball $B(x,\delta)$. Therefore $\operatorname{diam} f(C_{p_n(x)})<2\varepsilon$ and $\operatorname{diam}\operatorname{conv} f(C_{p_n(x)})<2\varepsilon$. Since $N_3\ge N_1$, $g_n^*(d)=f(d)$ for every vertex $d$ of the cube $C_{p_n(x)}\}$. Since $f_n^*(p_n(x))\in\operatorname{conv}\{g_n^*(d): d$ is a vertex of the cube $C_{p_n(x)}\}$, we see that $\|f_n^*(p_n(x))-f(p_n(x))\|<2\varepsilon$. Then 
$$\|f(x)-f_n(x)\|= \|f(x)-f(p_n(x))+ f(p_n(x))- f_n(p_n(x))+ f_n(p_n(x))-f_n(x)\|\le
\|f(x)-f(p_n(x))\|+ \|f(p_n(x))- f_n(p_n(x)) \|+ \|f_n(p_n(x))-f_n(x)\|<2\varepsilon+2\varepsilon+0.\square$$
Remark. Unfortunately, an extension of this construction fails for a non-separable Hilbert space $V$, because in this case there is no sequence of finitely-dimensional spaces with the dense in $V$ union. If we try to use the lattices $V_n=\{x\in V:n(x,e)\in\mathcal Z$ for each $e\in E\}$, where $E$ is a an orthonormal basis for the space $V$, the we shall have a problem with the affine extension from the lattice $V_n$ onto $V$. For instance, even if $V=\ell_2$, $n=1$ and $x=(1/n)$ then the counterparts of values $\lambda(d,x)$ for $d\in {0;1}^{\mathbb N}\cap\ell_2$ are equal to zero, because $\lim_{k\to\infty}(1-1/k)=0$. 
References 
[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989. 
