How to prove $\inf_{z\in\mathbb{R}^n}h(z)=\inf_{x\in{C}}f(x)$ for the function $h(z)=\min_{x\in{C}}\{f(x)+\frac{1}{2}||x-z||^2\}$ I have a problem related to convex optimization. I have a function where $h(z)=\min_{x\in{C}}\{f(x)+\frac{1}{2}||x-z||^2\}$ for all $z\in{\mathbb{R}^n}$. For this function, I want to show $\inf_{z\in\mathbb{R}^n}h(z)=\inf_{x\in{C}}f(x)$.
I tried to connect this problem to proximal point method because the format of the function $h(z)$ is the same as the form of the objective function in that method. By using proximal method, I mean that I tried to find relevant information in the steps of the proof of that method, but I did not have much of a success.
Another idea which I was told was to look at the problem in a different way, which I could not make a connection with the original problem and the suggested idea. The idea is about the fact that we can write a vector $y\in{\mathbb{R}}^n$ as $y=y_1+y_2$, where $y_1\in{\mathbb{R}^{n_1}}$ and $y_2\in{\mathbb{R}^{n_2}}$ and $n_1+n_2=n$. We can perform the minimization of a function with domain $y$ in two phases: first minimize over $y_1$, and then minimize over $y_2$.
I can not establish a clear connection between the idea explained above and the original problem.
If anyone can help with solving this problem, I would be very thankful!
 A: This is to supply some details at the request the OP to my comment to his/her question.

Suppose $\Phi:A\times B\rightarrow\mathbb{R}$ and that $\Phi$ is bounded below. Let
\begin{align}
\alpha&=\inf_{(x,y)\in A\times B}\Phi(x,y)\\
\beta&=\inf_{x\in A}\inf_{y\in B}\Phi(x,y)
\end{align}
The claim is that $\alpha=\beta$.
Notice that as $\Phi(x,y)\geq \alpha$ for all $(x,y)\in A\times B$. Then,  for $x'\in A$
$$\phi(x'):=\inf_{y\in B}\Phi(x',y)\geq \alpha$$
and so,
$$\beta=\inf_{x\in A}\phi(x)=\inf_{x\in A}\inf_{y\in B}\Phi(x,y)\geq \alpha$$
Conversely, for any $(x',y')\in A\times B$,
$$\Phi(x',y')\geq \phi(x')=\inf_{y\in B}\Phi(x',y)$$
Since $\phi(x')\geq \inf_{x\in A}\phi(x)$, we have that
$$\Phi(x',y')\geq \inf_{x\in A}\inf_{y\in B}\Phi(x,y)=\beta$$
It then follows that
$$\alpha=\inf_{(x,y)\in A\times B}\Phi(x,y)\geq \beta$$
by definition of $\inf$.
Putting things together, $\alpha=\beta$
A similar identity can be obtained by changing the order in the infirma, that is
$$\delta:=\inf_{y\in B}\inf_{x\in A}\Phi(x,y)=\inf_{(x,y)\in A\times B}\Phi(x,y)=\alpha$$
A: I think the solution is the following:
We can prove that a function in the form $h(z)=\min\limits_{x\in{C}}\{f(x)+\frac{1}{2}\|x-z\|^{2}\}$ for all $z\in\mathbb{R}^n$ has a unique solution for each $x\in{C}$, and attains its minimum.
We set $g(x)=f(x)+\frac{1}{2}\|x-z\|^{2}$. So, we can write:
$$h(z)=\min\limits_{x\in{C}}g(x)=\inf\limits_{x\in{C}}g(x),~\forall{z\in{\mathbb{R}^n}}$$
With optimality condition, we have:
$\langle{\nabla{f(x^*)}+\|x^*-z\|},x-x^*\rangle\geq{0}$.
Since $C\subseteq{\mathbb{R}^n}$, we can set $z=x^*$. (I missed to mention this in the question.)
$$\langle{\nabla{f(x^*)},x-x^*}\rangle\geq0$$
Thus, we have:
$$\inf\limits_{z\in\mathbb{R}^n}h(z)=\inf\limits_{x\in{C}}g(x)$$.
