What is the discontinuity set of the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$, $ f(x,y) := \sup \{ \sin (tx) + \sin (ty) : t \in \mathbb{R} \} ?$
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What have you tried? By computing that supremum, can you obtain a more manageable decription of the function? – Mariano Suárez-Álvarez May 13 '13 at 07:53
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It is clear that $f(0,y)=f(x,0)=1$ and $f(x,x)= 2.$ – user64494 May 13 '13 at 07:58
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You should also add $f(0,0) = 0$. $f(x,y) = 2$ when $x,y \neq 0$, no? I might be missing something.. – May 13 '13 at 08:15
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Why $f(x,y)=2$ if $x \neq 0$ and $y \neq 0$? – user64494 May 13 '13 at 08:34
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Have you considered $f(1,1/2)?$ – user64494 May 13 '13 at 09:11
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Clearly, $f(0,0)=0$, $f(0,y)=1$ for $y\ne0$, $f(x,0)=1$ for $x\ne0$.
If $x\ne0\ne y$ and $\alpha:=\frac xy$ is rational, say $\alpha=\frac nm$ with $n\in\mathbb Z, m\in\mathbb N$, $\gcd(n,m)=1$, then $$\tag1f(x,y)=\max_{t\in[0,2\pi]}\sin nt+\sin mt.$$ Especially, $f(x,-x)=0$. For all rationals $\alpha\ne -1$, the value $(1)$ is positive. If $n\equiv m\pmod 4$, the value in $(1)$ is $2$ and for all other rational $\alpha $, it is somewhere inbetween (because the maxima of $\sin nt$ don't match with those of $\sin mt$).
Finally, for irrational $\alpha$, one can readily see that $f(x,y)=2$.
By these results, $f$ is clearly discontinuos at all points $(x,y)$ with $x=0$ or $y=0$ or $\alpha=\frac nm$ rational with $n\not\equiv m\pmod 4$. At all other points, i.e. whenever $f(x,y)=2$, $f$ is continuous: For $\epsilon>0$, there exists $r>0$ such that $f_r(x,y):=\max\{\sin tx+\sin ty\mid -r\le t\le r\}>f(x,y)-\frac12\epsilon$. The function $f_r$ is continuous, hence $f(u,v)\ge f_r(u,v)>f(x,y)-\epsilon$ for $(u,v)$ close enough to $(x,y)$.
Hagen von Eitzen
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