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Given $f,g:\mathbb{R}\longrightarrow\mathbb{R}$ such that $f$ is not the zero function, and $\forall x\in\mathbb{R},\; |f(x)|\leq 1$, and $\forall x,y\in\mathbb{R},\; f(x+y)+f(x-y)=2f(x)g(y)$, can we claim $\forall y\in\mathbb{R},\; |g(y)|\leq 1$ ?... as it would be in the special case $f=\sin$ and $g=\cos$. Many thanks. Edit: title edited, I had forgotten the factor $2$.

P. Fazioli
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1 Answers1

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Let $M$ be the supremum of $|f(x)|$ on $\Bbb R$. For $0 < \epsilon < M$ choose $x \in \Bbb R$ such that $|f(x)| > M - \epsilon$. Then $$ |g(y)| = \frac{|f(x+y) + f(x-y)|}{2|f(x)|} \le \frac{M}{M - \epsilon} \, . $$ With $\epsilon \to 0$ it follows that $|g(y)| \le 1$.

The bound for $|g|$ is best possible, as the examples $f(x) = 1$, $g(y) = 1$, or $f(x) = \sin(x)$, $g(y) = \cos(y)$ show.

It is also necessary to require that $f$ is bounded, otherwise $f(x) = e^x$, $g(y) = (e^y + e^{-y})/2$ would be a counter-example.

Martin R
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  • One question: why One can choose choose $x$ with $0< \epsilon < M$ so that $|f(x)| > M - \epsilon $ . This extremely hard to grasp. can you explain why this true. – James Sep 21 '18 at 22:12
  • @JimmySabater: That follows from the definition of the supremum as the least upper bound. – Martin R Sep 21 '18 at 22:15
  • I thought the def of the supremum was that $|f(x) | \leq sup f $ – James Sep 21 '18 at 22:16
  • @JimmySabater: That would be an upper bound. The supremum is the least upper bound, which means that any smaller number (here: $M - \epsilon$) is not an upper bound. – Martin R Sep 21 '18 at 22:18
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    Im starting to understand and then the last inequality follows because of

    $$ |f(x+y) + f(x-y)| \leq |f(x+y)|+|f(x-y)| \leq M + M = 2 M $$

    is that the reason why?

     – James Sep 21 '18 at 22:20
  • But, then, I still have trouble seeing why we can let $\epsilon \to 0$. Why can we do that? – James Sep 21 '18 at 22:21
  • @JimmySabater: If $h(\epsilon) \ge c$ for all $\epsilon > 0$ and $\lim_{\epsilon \to 0} h(\epsilon)$ exists, then $\lim_{\epsilon \to 0} h(\epsilon) \ge c$. In this sense, taking limits preserves $\le$ and $\ge$ (provided that the limits exist). – Martin R Sep 21 '18 at 22:30
  • Thank you Martin R! – P. Fazioli Sep 22 '18 at 07:03