Let $X$ be a finite set and $f_1,\cdots f_k$ are disjointly supported function from $X$ to $\mathfrak{R}$, meaning that $\forall x \in X$ and $i \neq j$ both $f_i(x)$ and $f_j(x)$ can not be non zero simultaneously. In other words $f_i(x)f_j(x)=0 \forall x \in X , i\neq j$.
Show that $(f(x)-f(y))^2 \leq 2 \sum_{i=1}^k(f_i(x)-f_i(y))^2, \forall x,y\in X$, where $f=\sum_{i}f_i$.
My approach
$(f(x)-f(y))^2 = (\sum_i f_i(x) - \sum_i f_i(y))^2$
$=(\sum_if_i(x))^2 + (\sum_if_i(y))^2 - 2\sum_if_i(x)\sum_if(y)$
$=\sum_if_i(x)^2 + \sum_if_i(y)^2 -2\sum_if_i(x)f_i(y) - 2\sum_{i\neq j} f_i(x)f_j(y)$
$=\sum_i(f_i(x)-f_i(y))^2 - 2\sum_{i\neq j} f_i(x)f_j(y)$
In the above calculation of $(\sum_if_i(x))^2$, we have used the disjointedness property. Now I am kind of stuck. Can someone please figure it out? Thank you.
The above question is from here [page 160].
I would proceed as follows: If $$ f(x) = f_n(x) \,,\quad f(y) = f_n(x) $$ for some index $n$ then $$ \bigl(f(x) - f(y)\bigr)^2 = \bigl(f_n(x) - f_n(y)\bigr)^2 \le 2 \sum_{i=1}^k \bigl(f_i(x)-f_i(y) \bigr)^2 \, . $$ Otherwise there are indices $m \ne n$ such that $$ f(x) = f_m(x) \, , \quad f(y) = f_n(y) \, , \\ f_n(x) = 0 \, , \quad f_m(y) = 0 \, . $$ Then $$ \bigl(f(x) - f(y) \bigr)^2 = \bigl(f_m(x) - f_n(y) \bigr)^2 \\ \sum_{i=1}^k \bigl(f_i(x)-f_i(y) \bigr)^2 = f_m(x)^2 + f_n(y)^2 $$ and the inequality reduces to $$ \bigl(f_m(x) - f_n(y) \bigr)^2 \le 2 \bigl( f_m(x)^2 + f_n(y)^2 \bigr) $$ which is easily verified.