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Let $P$ be the set of all polynomials, with complex coefficients, in a variable $t$. For $x$ in $P$ the function $y$ is defined by $y\left( x\right) =\dfrac {d^{2}x} {dt^2}|_{t=1}$ Is $y$ a linear functional? Now I am aware of $y$ being a scalar valued function and the defining property of a linear functional is $$y\left( \alpha _{1}x_{1}+\alpha_{2}x_{2}\right) =\alpha _{1}y\left( x_{1}\right) +\alpha _{2}y\left( x_{2}\right)$$ It is the actual definition of $y$ which I am confused about.

Any help would be much appreciated.

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Why are Latex expression not showing up correctly ? –  Hardy Feb 13 '12 at 19:55
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Put math inside dollar ($) signs. Please make sure my edit didn't introduce errors. –  David Mitra Feb 13 '12 at 19:58
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Do you mean $dt^2$ instead of $dt^{-2}$? –  Abhishek Parab Feb 13 '12 at 20:01
    
As it stands now it is correct thanks for your help and the info. –  Hardy Feb 13 '12 at 22:02

1 Answer 1

up vote 1 down vote accepted

Keep in mind that $x(t)$ is a polynomial in the variable $t$. We have the map $y: x(t)\mapsto x''(1)$. The map takes a polynomial $x$ and returns a scalar value, the value given by the second derivative of $x$ evaluated at $t=1$. In order to see why this is linear, you have to check three things:

  • Differentiation is linear: $(\alpha \, x_1(t)+\beta\, x_2(t))'=\alpha x_1'(t)+\beta x_2'(t)$.
  • "Plugging in $t=a$" (here $a$ is fixed) is linear: $\big(\alpha x_1+\beta x_2\big)|_{t=a}=\alpha\big(x_1\big)|_{t=a}+\beta\big(x_2\big)|_{t=a}.$
  • The composition of two linear maps is itself linear.
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Thanks buddy u made it really simple i thought there might have been more to t = 1 but obviously not. –  Hardy Feb 13 '12 at 22:09

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