I try to solve the following problem:
Let $\varphi: D \to \mathbb{R}$ with $-\varphi(x) = \varphi(-x)$. Then $\underset{x \in D}{sup} \;(\varphi(x) - \varphi(\tilde{x}))\leq \underset{x\in D}{sup} \left(\varphi(x) - \underset{\tilde{x} \in D}{inf} \; \varphi(\tilde{x})\right)$. So my idea behind this inequality is that the difference only gets bigger if I subtract the smallest function value. This should be okay, I think.
I want to obtain the following inequality: $\underset{x \in D}{sup} \;(\varphi(x) - \varphi(\tilde{x})) \leq \underset{x \in D}{sup} \; \underset{\tilde{x} \in D}{inf} \left(\varphi(x) - \varphi(\tilde{x})\right)$.
So my idea is to use the first inequality and move on with: $\begin{align*} \underset{x \in D}{sup} \;(\varphi(x) - \varphi(\tilde{x}))&\leq \underset{x\in D}{sup} \left(\varphi(x) - \underset{\tilde{x} \in D}{inf} \; \varphi(\tilde{x})\right) &\leq \underset{x \in D}{sup} \; \underset{\tilde{x} \in D}{inf} \left(\varphi(x) - \varphi(\tilde{x})\right), \end{align*}$
but I'm really not sure if this holds. I think it could be because of $-\varphi(x) = \varphi(-x)$. So basically my question is if $\varphi(x) - \underset{\tilde{x}\in D }{inf} \varphi(\tilde{x}) = \underset{\tilde{x} \in D}{inf} \left(\varphi(x) - \varphi(\tilde{x})\right)$ holds. Can someone please help me with figuring this out?
The equality
$$\inf_{\tilde{x}\in D}\left(\varphi(x)-\varphi(\tilde{x})\right)\color{red}=\varphi(x)-\inf_{\tilde{x}\in D}\varphi(\tilde{x})$$
does not hold for all such $\varphi$. As a counterexample, consider $\varphi=\sin:\mathbb{R}\to\mathbb{R}$. Then
$$-\sin(x)=\sin(-x)$$
for all $x\in\mathbb{R}$, but
$$\inf_{\tilde{x}\in\mathbb{R}}\left(\sin(x)-\sin(\tilde{x})\right)=\sin(x)-1,$$
but
$$\sin(x)-\inf_{\tilde{x}\in\mathbb{R}}\sin(\tilde{x})=\sin(x)+1.$$