Prove a function realizes a maximum.

[sleek juniper]

Let $\mathcal{F}C$ be the set of Lipschitz functions that integrate to 1 with Lipschitz constant at most $C$. Let $x_1, \cdots, x_n \in \mathbb{R}$ and $\mathcal{L}n(f) = \prod{i=1}^n f(x_i)$. Show that $\mathcal{L}n$ reaches a maximum in $\mathcal{F}C$.\
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There is a hint that says that I should first find a candidate and then prove that no function can surpass it. My candidate is the following:\
Define $g{y_1, \cdots, y_n}(x)=\max{1 \leq i \leq n} (y_i - C \cdot d(x,x_i))^+$.\
That function is the maximum of some cones of slope $C$ and maximum value of $y_i$ at $x_i$, and those cones are truncated so that they aren't negative.\
Now the following holds: $\mathcal{L}n(g{y_1, \cdots, y_n})=\prod{i=1}^n y_i$.\
Let $I(y_1, \cdots, y_n) = \int_{\mathbb{R}} g_{y_1, \cdots, y_n}$ be a continuous function. It's not hard to see that $I^{-1}(1)$ is compact, so $\mathcal{L}n(g{y_1, \cdots, y_n})$ reaches a maximum in $I^{-1}(1)$. Let's call $(z_1, \cdots, z_n)$ the point that realizes the maximum.\
So now $g_{z_1, \cdots, z_n}$ should be the global maximum, but how can I prove that?

[muted junco]

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[worldly spoke]

Casiel368

[sleek juniper]

Please ping if someone answers

[sly quarry]

@sleek juniper does the ^+ on the top resemble max{ * , 0}?

[sleek juniper]

Yes

[sly quarry]

@sleek juniper did you do this yet?

ill give it a go