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This question is inspired by this question. Is it always true that if $\{f_n(x)\}_n$ are linearly independent then so is $\{f_n(x)+f_n(-x)\}$ given that each $f_n(x)$ do not have definite parity?


2 Answers 2

The condition "do not have definite parity" is not strong enough. For example, look at $1+x$, $1+x^3$, $1+x^5$, and so on. In each case, when we calculate $f(x)+f(-x)$, we get $2$.


No. Consider $\{f_1,f_2\}$ with $f_2(x)=f_1(-x)$ for every $x$. Let $g_n(x)=f_n(x)+f_n(-x)$. Then $g_1=g_2$ hence $\{g_1,g_2\}$ is dependent. If $\{f_1,f_2\}$ is linearly independent, this is a counterexample. To get linearly independent $\{f_1,f_2\}$, choose any $f_1$ such that $f_1(x)$ is positive for the positive values of $x$ and zero for the negative values of $x$.


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