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I am not being able to find a clear answer to the following question with my limited knowledge of algebra.

Let us consider $n$ real-valued functions {$f_{i}(t): t \in \mathbb{R}$ and $ i={1,2,3,...,n}$}.
If any of {$f_{i}(t)$} are discontinuous at $t=a$, can all the nontrivial elementary symmetric polynomials made with {$f_{i}(t)$} be continuous at $t=a$?

Probably the answer is negative. Can anybody please help find a definitive but easy-to-understand answer to this (Preferably using analysis)?

Update: 2016 Jan 08

As fellow stackexchangers promptly pointed out, permuting the values of {$f_{i}(t)$} at $t=a$ leaves the symmetric polynomials intact. I must add that permutation of {$f_{i}(t)$} is not allowed to create discontinuities at $t=a$.

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sobasu
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    Not sure what you mean by polynomials "made with ${f_i(t)}$". Made with? How? – MPW Jan 08 '16 at 13:15
  • @MPW: By composition, I guess: $e_k(f_1\ldots f_n)$, where $e_k$ is the $k$-th symmetric polynomial in $n$ variables. This is an interesting question, whose proof is probably a brute-force check. (I agree that the answer should be negative) – Giuseppe Negro Jan 08 '16 at 13:17
  • Let's tackle the simplest cases, say the first degree elementary symmetric polynomial, or the case $n=2$. – hardmath Jan 08 '16 at 13:18
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    I considered the simplest case and it gives a negative answer. So clearly it cannot be true for all $n \in \mathbb{N}$. But can it be true for some $n$? – sobasu Jan 08 '16 at 13:25
  • @sobasu: I'm sure the problem is clear to you, but if you would illustrate what is meant by "the answer is negative" in a simple case, it would help clear up the meaning for your Readers. E.g. are we trying to pick a pair of discontinuous functions $f_1(t),f_2(t)$ such that both $f_1(t)+f_2(t)$ and $f_1(t)\cdot f_2(t)$ are continuous? – hardmath Jan 08 '16 at 13:42
  • @hardmath: Thanks. For $n=2$, it means exactly what you said. – sobasu Jan 08 '16 at 13:47
  • In the case $n=2$ it is possible. Take for example $f_1(x)=\begin{cases}1\quad x<1\ 2\quad x≥1\end{cases}$, $f_2(x)=\begin{cases}2\quad x<1\ 1\quad x≥1\end{cases}$. Then $f_1+f_2=3$ and $f_1 \cdot f_2 = 2$ are both constant. – s.harp Jan 08 '16 at 13:53
  • general answer can follow inductively – s.harp Jan 08 '16 at 13:57
  • @s.harp: I was just calculating something in this fashion. If $f_{1}(x)$ and $f_{2}(x)$ has jump discontinuities of amount $2 \lim_{x\to a-} f_{2}(x)$ and $2 \lim_{x\to a-} f_{1}(x)$ and also satisfies $\lim_{x\to a-} f_{1}(x)+f_{2}(x)=0$, it can hold. – sobasu Jan 08 '16 at 13:59

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Let $f_1(t) = 1$ for $t\gt 0$ and equal $0$ otherwise. Of course $f_1(t)$ is discontinuous at $t=0$, and so is $f_2(t) = 1-f_1(t)$.

But both "elementary symmetric polynomial" compositions $f_1(t) + f_2(t) = 1$ and $f_1(t)\cdot f_2(t) = 0$ are constant functions and everywhere continuous.

We can extend $n=2$ to larger values by throwing in zero functions $f_3(t),\ldots$.

hardmath
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    It is actually enough that the functions $f_n$ for $n>2$ be continuous, because then $\sum_{1≤i_1<..<i_k≤n}f_{i_1}..f_{i_k} = \sum_{1≤i_1<..<i_k≤n-1}f_{i_1}..f_{i_k} + \left(\sum_{1≤i_1<..<i_{k-1}≤n-1}f_{i_1}..f_{i_{k-1}}\right) f_n$ will be continuous. – s.harp Jan 08 '16 at 14:00
  • @ s.harp and hardmath: Thanks for the nice examples. Looking at those, I thought that is we look at the problem using coordinate geometry, then we can look at the simultaneous solutions for $e_{k}(f1…fn)=c_{k}$ and permute the solutions at the discontinuity without changing the symmetric polynomials. – sobasu Jan 08 '16 at 14:10
  • I just added the motivation of my question in the original post. "The motivation of this problem is to show that the roots of a polynomial of degree $n$ vary continuously when the coordinates are varied continuously, in the range of the coordinates for which the polynomial has $n$ real solutions." Maybe my question needs more precise rephrasing for this. – sobasu Jan 08 '16 at 14:14
  • @s.harp and hardmath: While considering the roots and coordinates of polynomials, permuting the roots does not change the polynomial. Since all the examples we fabricated involve permutation of the values at the point of discontinuity, it does not answer the question about polynomials. What happens if permutations are forbidden? Can you think of a better rephrasing of the original question in that case? – sobasu Jan 08 '16 at 14:27
  • Your original question had to do with continuity of elementary symmetric polynomials evaluated at $f_1(t),\ldots,f_n(t)$. In that context you have no way to properly "forbid" permutation of roots (of a degree $n$ polynomial) because those roots do not appear in the problem formulation. If you are interested in the proof that polynomial roots are continuously dependent on the coefficients, I would check to see if that Question has not already been answered here. – hardmath Jan 08 '16 at 19:35
  • @ Hardmath. Thanks. I put an update to restrict permutation. The theory of symmetric groups already answers this question . I am aware of complex analytic proofs of continuous dependence of coefficients and roots. My motivation is to prove the fundamental theorem of algebra for real polynomials without assuming the existence of complex numbers. I am avoiding Jordan's curve theorem and the only part left is the question I asked with permutations restricted. – sobasu Jan 08 '16 at 22:46