What does it mean if a function $f(x)=f(2\alpha-x)$?

What does it mean if a function $f(x)=f(2\alpha-x)$?

Question

In this question we fix a real number $\alpha$ which will be the same throughout. We say that a function $f$ is bilateral if $f(x) = f(2\alpha - x)$ for all $x$.

1. Show that if $f(x) = (x-\alpha)^2$ for all $x$ then the function $f$ is bilateral.

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1. On the other hand, if $f(x) = x-\alpha$ for all $x$ then the function $f$ is not bilateral.

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1. Show that if $n$ is a non-negative integer and $a$ and $b$ are any real numbers then $\int_a^b x^n \:dx = -\int_b^a x^n \:dx.$

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1. Hence show that if $f$ is a polynomial (and $a$ and $b$ are any reals) then $\int_a^b f(x) \:dx = -\int_b^a f(x) \:dx.$

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1. Suppose that $f$ is any bilateral function. By considering the area under the curve $y = f(x)$ explain why for any $t \geq \alpha$ we have $\int_\alpha^t f(x) \:dx = \int_{2\alpha-t}^\alpha f(x) \:dx.$

If $f$ is a function we write $G$ for the function defined by $G(t) = \int_\alpha^t f(x) \:dx$ for all $t$.

1. Suppose now that $f$ is any bilateral function. Show that $G(t) = -G(2\alpha-t)$ for all $t$.

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1. Suppose that $f$ is a bilateral polynomial such that $G$ is also bilateral. Show that $G(x) = 0$ for all $x$.

Solution

We have $f(x) = f(2\alpha - x) \iff f((x-\alpha)+\alpha)= f(-(x-\alpha)+\alpha),$ and so $f(x)$ is bilateral if and only if $x=\alpha$ is a line of symmetry for the curve $y=f(x)$.

In the special case of $\alpha=0$, this would mean that $f$ is an even function.

If $f(x) = (x-\alpha)^2$, then $f(2\alpha - x) = ((2\alpha - x) -\alpha)^2 = (\alpha-x)^2 = (x-\alpha)^2 = f(x),$ for all $x$, and so $f$ is bilateral.

If $f(x) = x-\alpha$, then $f(2\alpha - x) = ((2\alpha - x) -\alpha) = (\alpha-x) = -f(x),$ for all $x$, and so $f$ is not bilateral.

The only solution to $f(x) = -f(x)$ for all $x$ is $f(x) = 0$ for all $x$, which is not the $f(x)$ we have here.

We know $\displaystyle\int_a^b x^n \:dx = \left[\dfrac{x^{n+1}}{n+1}\right]_a^b = \dfrac{b^{n+1}-a^{n+1}}{n+1}$.

We also know that $\displaystyle\int_b^a x^n \:dx = \left[\dfrac{x^{n+1}}{n+1}\right]_b^a = \dfrac{a^{n+1}-b^{n+1}}{n+1}$.

Thus we have that $\displaystyle\int_a^b x^n \:dx = -\displaystyle\int_b^a x^n \:dx$.

Let $f(x)$ be the polynomial $\sum\limits_{0}^{n} a_nx^n$. Then

$$\begin{align*} \int_a^b f(x) \:dx &= \int_a^b \sum_0^n a_nx^n \:dx \\ &= \sum_0^n \left(a_n\left(\int_a^b x^n \:dx\right)\right)\\ &= \sum_0^n \left(a_n\left(-\int_b^a x^n \:dx\right)\right)\\ &= -\int_b^a \sum_0^n a_nx^n \:dx\\ &= -\int_b^a f(x) \:dx. \end{align*}$$

1. Suppose that $f$ is any bilateral function. By considering the area under the curve $y = f(x)$ explain why for any $t \geq \alpha$ we have $\int_\alpha^t f(x) dx = \int_{2\alpha-t}^\alpha f(x) dx.$

The symmetry of the bilateral function $f(x)$ about $x = \alpha$ guarantees that the two areas in the diagram above are equal, and so $\int_\alpha^t f(x) \:dx = \int_{2\alpha-t}^\alpha f(x) \:dx.$

Using the results from above we have

$$\begin{align*} -G(2\alpha-t) &= -\int_\alpha^{2\alpha-t} f(x) \:dx\\ &= \int_{2\alpha-t}^\alpha f(x) \:dx\\ &= \int_{\alpha}^t f(x) \:dx\\ &= G(t). \end{align*}$$

If $f$ is bilateral, then $G(t) = -G(2\alpha-t)$, but if $G$ is bilateral, then $G(2\alpha-t)=G(t)$.

Thus on adding these equations $G(t) = 0$ for all $t$ (or equivalently $G(x) = 0$ for all $x$).