# Is $y\left( x\right) =\dfrac {d^{2}x} {dt^2}|_{t=1}$ a linear functional for vector space of polynomials?

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
Put math inside dollar ($) signs. Please make sure my edit didn't introduce errors. – David Mitra Feb 13 '12 at 19:58 Do you mean$dt^2$instead of$dt^{-2}$? – Abhishek Parab Feb 13 '12 at 20:01 As it stands now it is correct thanks for your help and the info. – Hardy Feb 13 '12 at 22:02 ## 1 Answer Keep in mind that$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