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yes, it is the only case. Denote $g(x)=(\int{|f(x,y)|^pd\nu(y))}^{1/p}$, then your condition is $||g||_1=||g||_p$ which is known to imply that $g$ is a constant function, see herehere.
yes, it is the only case. Denote $g(x)=(\int{|f(x,y)|^pd\nu(y))}^{1/p}$, then your condition is $||g||_1=||g||_p$ which is known to imply that $g$ is a constant function, see here.
yes, it is the only case. Denote $g(x)=(\int{|f(x,y)|^pd\nu(y))}^{1/p}$, then your condition is $||g||_1=||g||_p$ which is known to imply that $g$ is a constant function, see here.
yes, it is the only case. Denote $g(x)=(\int{|f(x,y)|^pd\nu(y))}^{1/p}$, then your condition is $||g||_1=||g||_p$ which is known to imply that $g$ is a constant function, see here.
