Is There a Limit?

Dan Sitaru and Leo Giugiuc posed and solved the following problem involving a functional equation:

Let $f: \mathbb{R}\rightarrow [0,2]$ be a function for which, for all $x,$ $f^{2}(x-2)+f^{2}(x+2)=3.$ Study the existence of the limit $\displaystyle\lim_{x\rightarrow\infty}f(x).$

Solution

Substitute $x+6$ for $x$:

$f^{2}(x+4)+f^{2}(x+8)=3.$

Substitute $x+2$ for $x$:

$f^{2}(x)+f^{2}(x+4)=3.$

The two equations imply $f^{2}(x+8)=f^{2}(x),$ and, since $f$ is positive valued, $f(x+8)=f(x),$ meaning that $f$ is a periodic function. A periodic function has a limit at infinity only if it is constant. If it is, $2f^{2}(x)=3$ so that $f(x)=\sqrt{3/2}$ which is also its limit at infinity. Otherwise, the limit does not exist.

Critical remarks

Several pieces of data in the problem provide no useful information, nor are specifically helpful in solving the problem. They may only serve to distract from the essence of the problem. I refer to such occurences as a red herring. The function could have been defined as $f: \mathbb{R}\rightarrow\mathbb{R}^{+},$ just a positive real valued function. That the two squares add up to $3$ is of no consequence, except for the obtaining the value of the function in case it's constant. The same holds for the two values of the function being squared and for the two values being taken at $x-2$ and $x+2.$ Here's a simplified formulation:

Let $h: \mathbb{R}\rightarrow\mathbb{R}^{+}$ be a function for which, for all $x,$ $h(x)+h(x+a)=b,$ where $a$ and $b$ are constant. Study the existence of the limit $\displaystyle\lim_{x\rightarrow\infty}h(x).$

Solution

Substitute $x-a$ for $x$ to obtain $h(x-a)+h(x)=b,$ from which $h(x-a)=h(x+a),$ implying a period of $2h$ and, as a consequence, no limit at infinity unless $h$ is a constant. In that case, the constant value of the function and its limit is $b/2.$

For the solution of the original problem take $h(x)=f^{2}(x-2),$ $a=4,$ $b=3.$

(The problem is related to Problem 5, 1968 IMO.)

What Is Red Herring

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