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Let $f:2^V\to \mathbb{R}$ be a symmetric submodular function. Let $T\subset V$, consider the function $f_T:2^T\to \mathbb{R}$ defined as follows: $$f_T(X) = \min \{f(Y) | X\subset Y,T\backslash X\subset V\backslash Y \}$$

Is $f_T$ submodular? If so, do we need $f$ to be symmetric? It be best if there is a known reference.

I can prove each condition considered independently is submodular when $f$ is submodular. say $g_T,\overline{g_T}:2^T\to \mathbb{R}$ defined as $$g_T(X) = \min \{f(Y) | X\subset Y\}$$ $$\overline{g_T}(X) = \min \{f(Y) | T\backslash X\subset V\backslash Y\}$$

Lemma 1: Let $$f_*(X) = \min \{ f(Y) : Y\subset X\}$$ and $$f^*(X) = \min \{ f(Y) : X \subset Y\}$$. If $f$ is submodular then $f_*$ and $f^*$ are submodular.

This can be proven easily from definition, and one can find reference for $f_*$ as proposition 2.1 in Submodular functions and convexity.

If $f:2^V \to \mathbb{R}$, then define $f':2^V \to \mathbb{R}$ as $f'(X)=f(V\backslash X)$.

Lemma 2: $f$ is submodular if and only if $f'$ is submodular.

$g_T$ is submodular because $g_T$ is just $f^*$ restricted on $T$.

$\overline{g_T}$ is submodular \begin{align*} (\overline{g_T})'(X) &= \overline{g_T}(T\backslash X)\\ &=\min \{f(Y) | T\backslash (T\backslash X)\subset V\backslash Y\}\\ &=\min \{f(Y) | X\subset V\backslash Y\}\\ &=\min \{f(Y) | Y\subset V\backslash X\}\\ &= (f_*)'(X) \\ \end{align*}

So $(\overline{g_T})'$ is $(f_*)'$ restricted to $T$, thus $\overline{g_T}$ is submodular.

Chao Xu
  • 5,768

1 Answers1

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Yes it is, and the symmetry is not required. Let $C_T(X) = \{Y|X\subset Y, T\setminus X\subset V\setminus Y\}$ Let $X^* = argmin \{f(Y)|C_T(X)\}$.

One can see that $X\subset X^*$ and $T\setminus X\subset V\setminus X^*$.

Thus we have $A\cup B\subset A^*\cup B^*$ and $T\setminus (A\cup B) \subset V\setminus (A^*\cup B^*)$, thus $A^*\cup B^*\in C_T(A\cup B)$, and we have $f(A^*\cup B^*)\geq f((A\cup B)^*)$.

Similarly, we have $f(A^*\cap B^*)\geq f((A\cap B)^*)$.

\begin{align*} f_T(A)+f_T(B) &= f(A^*)+f(B^*)\\ &\geq f(A^*\cup B^*)+f(A^*\cap B^*)\\ &\geq f((A\cup B)^*)+f((A\cap B)^*)\\ &= f_T(A\cup B)+f_T(A\cap B) \end{align*}

Chao Xu
  • 5,768