If two functions are linearly dependent within some interval $I$ then is it possible for them to be linearly independent on some interval $J$ which is contained within $I$? If yes then please help me find an example.
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Yes, it is possible. Consider the functions
$$ f(x) = \chi_{(0,1)}(0) = \begin{cases} 1 & \text{if $0 < x < 1$}, \\ 0 & \text{otherwise}. \end{cases} $$
and
$$ g(x) = \chi_{(2,3)}(x) = \begin{cases} 1 & \text{if $2 < x < 3$}, \\ 0 & \text{otherwise}. \end{cases} $$
$f$ is clearly not a scalar multiple of $g$ on the interval $(0,3)$ (since $1 \neq c \cdot 0$ for any $c$), but $f = g$ on the interval $(1,2)$.
mephistolotl
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This concept can be readily extended to pretty much any “niceness” condition you wish. In general, the value of $f$ on $(a,b)$ doesn’t tell you anything about the value of $f$ on $(c,d)$ for $b<c$. – Stella Biderman Dec 01 '17 at 06:28
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@StellaBiderman I agree that you can require any degree of smoothness and use a bump function. However, I don't believe it's possible if you require analyticity, since you can reconstruct an analytic function from any open interval. So I think there is a limit to how nice such functions can be. – mephistolotl Dec 01 '17 at 06:37
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@mephistolotl Hi, I was wondering why f and g are linearly independent on (0,3). Because I thought they are linearly dependent since f is 0 on [1,3) and g is 0 on (0,2] which means they are zero vectors on the intervals and so they are linearly dependent. Are they linearly independent since they are not the same as zero vector on (0,3)? Sorry for this random question. – whwjddnjs Sep 27 '19 at 09:46
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@JungWonCho Try to find two constants, $a$ and $b$, such that $af+bg=0$ on $(0,3)$. For $x=0.5$ you have $a\cdot f(0.5) + b\cdot g(0.5) = a = 0$ and for $x=2.5$ you have $a\cdot f(2.5) + b\cdot g(2.5) = b = 0$. Therefore ${f, g}$ is linearly independent if the domain is $(0,3)$. – mephistolotl Sep 28 '19 at 03:54