If $f$ is $mathbb R^2$-differentiable and the limit $limlimits_{h to 0} Releft(frac{f(z+h)-f(z)}{h}right)$...












1












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Assume $f$ is defined on some domain $Dsubsetmathbb C$, differentiable on $D$ as a function of two real variables, and that, for all $zin D$, the limit $limlimits_{h to 0} Releft(frac{f(z+h)-f(z)}{h}right)$ exists and is finite. Does this imply that $f$ is holomorphic on $D$?



This was a task on my exam today, which I didn't manage.






complex-analysis analytic-functions







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    @MisterRiemann That example depends on if $hto 0$ has $h$ on the real line or complex. If $h$ is complex, that would reduce to $\lim_{hto . 0}frac{overline h}{h}=c_z,$ but this limit doesn't converge.
    $endgroup$
    – Thomas Andrews
    Nov 29 '18 at 20:11










  • $begingroup$
    @ThomasAndrews Good point, thanks for the remark. I shall remove the comment.
    $endgroup$
    – MisterRiemann
    Nov 29 '18 at 20:12












  • $begingroup$
    h is complex of course
    $endgroup$
    – ryszard eggink
    Nov 29 '18 at 20:13
















1












$begingroup$


Assume $f$ is defined on some domain $Dsubsetmathbb C$, differentiable on $D$ as a function of two real variables, and that, for all $zin D$, the limit $limlimits_{h to 0} Releft(frac{f(z+h)-f(z)}{h}right)$ exists and is finite. Does this imply that $f$ is holomorphic on $D$?



This was a task on my exam today, which I didn't manage.






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @MisterRiemann That example depends on if $hto 0$ has $h$ on the real line or complex. If $h$ is complex, that would reduce to $\lim_{hto . 0}frac{overline h}{h}=c_z,$ but this limit doesn't converge.
    $endgroup$
    – Thomas Andrews
    Nov 29 '18 at 20:11










  • $begingroup$
    @ThomasAndrews Good point, thanks for the remark. I shall remove the comment.
    $endgroup$
    – MisterRiemann
    Nov 29 '18 at 20:12












  • $begingroup$
    h is complex of course
    $endgroup$
    – ryszard eggink
    Nov 29 '18 at 20:13














1












1








1





$begingroup$


Assume $f$ is defined on some domain $Dsubsetmathbb C$, differentiable on $D$ as a function of two real variables, and that, for all $zin D$, the limit $limlimits_{h to 0} Releft(frac{f(z+h)-f(z)}{h}right)$ exists and is finite. Does this imply that $f$ is holomorphic on $D$?



This was a task on my exam today, which I didn't manage.






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$




Assume $f$ is defined on some domain $Dsubsetmathbb C$, differentiable on $D$ as a function of two real variables, and that, for all $zin D$, the limit $limlimits_{h to 0} Releft(frac{f(z+h)-f(z)}{h}right)$ exists and is finite. Does this imply that $f$ is holomorphic on $D$?



This was a task on my exam today, which I didn't manage.






complex-analysis analytic-functions




complex-analysis analytic-functions






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Nov 29 '18 at 21:42









Did

247k23222457




247k23222457










asked Nov 29 '18 at 20:06









ryszard egginkryszard eggink

308110




308110








  • 1




    $begingroup$
    @MisterRiemann That example depends on if $hto 0$ has $h$ on the real line or complex. If $h$ is complex, that would reduce to $\lim_{hto . 0}frac{overline h}{h}=c_z,$ but this limit doesn't converge.
    $endgroup$
    – Thomas Andrews
    Nov 29 '18 at 20:11










  • $begingroup$
    @ThomasAndrews Good point, thanks for the remark. I shall remove the comment.
    $endgroup$
    – MisterRiemann
    Nov 29 '18 at 20:12












  • $begingroup$
    h is complex of course
    $endgroup$
    – ryszard eggink
    Nov 29 '18 at 20:13














  • 1




    $begingroup$
    @MisterRiemann That example depends on if $hto 0$ has $h$ on the real line or complex. If $h$ is complex, that would reduce to $\lim_{hto . 0}frac{overline h}{h}=c_z,$ but this limit doesn't converge.
    $endgroup$
    – Thomas Andrews
    Nov 29 '18 at 20:11










  • $begingroup$
    @ThomasAndrews Good point, thanks for the remark. I shall remove the comment.
    $endgroup$
    – MisterRiemann
    Nov 29 '18 at 20:12












  • $begingroup$
    h is complex of course
    $endgroup$
    – ryszard eggink
    Nov 29 '18 at 20:13








1




1




$begingroup$
@MisterRiemann That example depends on if $hto 0$ has $h$ on the real line or complex. If $h$ is complex, that would reduce to $\lim_{hto . 0}frac{overline h}{h}=c_z,$ but this limit doesn't converge.
$endgroup$
– Thomas Andrews
Nov 29 '18 at 20:11




$begingroup$
@MisterRiemann That example depends on if $hto 0$ has $h$ on the real line or complex. If $h$ is complex, that would reduce to $\lim_{hto . 0}frac{overline h}{h}=c_z,$ but this limit doesn't converge.
$endgroup$
– Thomas Andrews
Nov 29 '18 at 20:11












$begingroup$
@ThomasAndrews Good point, thanks for the remark. I shall remove the comment.
$endgroup$
– MisterRiemann
Nov 29 '18 at 20:12






$begingroup$
@ThomasAndrews Good point, thanks for the remark. I shall remove the comment.
$endgroup$
– MisterRiemann
Nov 29 '18 at 20:12














$begingroup$
h is complex of course
$endgroup$
– ryszard eggink
Nov 29 '18 at 20:13




$begingroup$
h is complex of course
$endgroup$
– ryszard eggink
Nov 29 '18 at 20:13










1 Answer
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Let us fix some $z$ in $D$ and use the notations $$h=r+isqquad f_x(z)=a+ibqquad f_y(z)=c+id$$ for some real numbers $r$, $s$, $a$, $b$, $c$ and $d$. Then the $mathbb R^2$-differentiability of the function $f$ at $z$ reads, in the limit $hto0$ or, equivalently, in the limit $(r,s)to(0,0)$, $$f(z+h)-f(z)=(a+ib)r+(c+id)s+o(|h|)$$
that is,
$$frac{f(z+h)-f(z)}h=frac{(f(z+h)-f(z))bar h}{|h|^2}=frac{((a+ib)r+(c+id)s)(r-is)}{r^2+s^2}+o(1)$$
Thus, the hypothesis is exactly the existence of the limit
$$lim_{(r,s)to(0,0)}frac{ar^2+(b+c)rs+ds^2}{r^2+s^2}$$ Considering the limits of this ratio when $rto0$ along the lines $s=rt$, for each $t$ fixed, one sees that $$frac{a+(b+c)t+dt^2}{1+t^2}$$ should be independent of $t$. In particular, the cases $t=0$ and $ttoinfty$ yield the condition $$a=d$$ and the cases $t=pm1$ yield the condition $$b+c=0$$ Since these are Cauchy-Riemann equations for $f$ at $z$, the proof is complete.






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  • 1




    $begingroup$
    (+1) Clear and concise. Well written.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:41











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$begingroup$

Let us fix some $z$ in $D$ and use the notations $$h=r+isqquad f_x(z)=a+ibqquad f_y(z)=c+id$$ for some real numbers $r$, $s$, $a$, $b$, $c$ and $d$. Then the $mathbb R^2$-differentiability of the function $f$ at $z$ reads, in the limit $hto0$ or, equivalently, in the limit $(r,s)to(0,0)$, $$f(z+h)-f(z)=(a+ib)r+(c+id)s+o(|h|)$$
that is,
$$frac{f(z+h)-f(z)}h=frac{(f(z+h)-f(z))bar h}{|h|^2}=frac{((a+ib)r+(c+id)s)(r-is)}{r^2+s^2}+o(1)$$
Thus, the hypothesis is exactly the existence of the limit
$$lim_{(r,s)to(0,0)}frac{ar^2+(b+c)rs+ds^2}{r^2+s^2}$$ Considering the limits of this ratio when $rto0$ along the lines $s=rt$, for each $t$ fixed, one sees that $$frac{a+(b+c)t+dt^2}{1+t^2}$$ should be independent of $t$. In particular, the cases $t=0$ and $ttoinfty$ yield the condition $$a=d$$ and the cases $t=pm1$ yield the condition $$b+c=0$$ Since these are Cauchy-Riemann equations for $f$ at $z$, the proof is complete.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    (+1) Clear and concise. Well written.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:41
















4












$begingroup$

Let us fix some $z$ in $D$ and use the notations $$h=r+isqquad f_x(z)=a+ibqquad f_y(z)=c+id$$ for some real numbers $r$, $s$, $a$, $b$, $c$ and $d$. Then the $mathbb R^2$-differentiability of the function $f$ at $z$ reads, in the limit $hto0$ or, equivalently, in the limit $(r,s)to(0,0)$, $$f(z+h)-f(z)=(a+ib)r+(c+id)s+o(|h|)$$
that is,
$$frac{f(z+h)-f(z)}h=frac{(f(z+h)-f(z))bar h}{|h|^2}=frac{((a+ib)r+(c+id)s)(r-is)}{r^2+s^2}+o(1)$$
Thus, the hypothesis is exactly the existence of the limit
$$lim_{(r,s)to(0,0)}frac{ar^2+(b+c)rs+ds^2}{r^2+s^2}$$ Considering the limits of this ratio when $rto0$ along the lines $s=rt$, for each $t$ fixed, one sees that $$frac{a+(b+c)t+dt^2}{1+t^2}$$ should be independent of $t$. In particular, the cases $t=0$ and $ttoinfty$ yield the condition $$a=d$$ and the cases $t=pm1$ yield the condition $$b+c=0$$ Since these are Cauchy-Riemann equations for $f$ at $z$, the proof is complete.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    (+1) Clear and concise. Well written.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:41














4












4








4





$begingroup$

Let us fix some $z$ in $D$ and use the notations $$h=r+isqquad f_x(z)=a+ibqquad f_y(z)=c+id$$ for some real numbers $r$, $s$, $a$, $b$, $c$ and $d$. Then the $mathbb R^2$-differentiability of the function $f$ at $z$ reads, in the limit $hto0$ or, equivalently, in the limit $(r,s)to(0,0)$, $$f(z+h)-f(z)=(a+ib)r+(c+id)s+o(|h|)$$
that is,
$$frac{f(z+h)-f(z)}h=frac{(f(z+h)-f(z))bar h}{|h|^2}=frac{((a+ib)r+(c+id)s)(r-is)}{r^2+s^2}+o(1)$$
Thus, the hypothesis is exactly the existence of the limit
$$lim_{(r,s)to(0,0)}frac{ar^2+(b+c)rs+ds^2}{r^2+s^2}$$ Considering the limits of this ratio when $rto0$ along the lines $s=rt$, for each $t$ fixed, one sees that $$frac{a+(b+c)t+dt^2}{1+t^2}$$ should be independent of $t$. In particular, the cases $t=0$ and $ttoinfty$ yield the condition $$a=d$$ and the cases $t=pm1$ yield the condition $$b+c=0$$ Since these are Cauchy-Riemann equations for $f$ at $z$, the proof is complete.






share|cite|improve this answer











$endgroup$



Let us fix some $z$ in $D$ and use the notations $$h=r+isqquad f_x(z)=a+ibqquad f_y(z)=c+id$$ for some real numbers $r$, $s$, $a$, $b$, $c$ and $d$. Then the $mathbb R^2$-differentiability of the function $f$ at $z$ reads, in the limit $hto0$ or, equivalently, in the limit $(r,s)to(0,0)$, $$f(z+h)-f(z)=(a+ib)r+(c+id)s+o(|h|)$$
that is,
$$frac{f(z+h)-f(z)}h=frac{(f(z+h)-f(z))bar h}{|h|^2}=frac{((a+ib)r+(c+id)s)(r-is)}{r^2+s^2}+o(1)$$
Thus, the hypothesis is exactly the existence of the limit
$$lim_{(r,s)to(0,0)}frac{ar^2+(b+c)rs+ds^2}{r^2+s^2}$$ Considering the limits of this ratio when $rto0$ along the lines $s=rt$, for each $t$ fixed, one sees that $$frac{a+(b+c)t+dt^2}{1+t^2}$$ should be independent of $t$. In particular, the cases $t=0$ and $ttoinfty$ yield the condition $$a=d$$ and the cases $t=pm1$ yield the condition $$b+c=0$$ Since these are Cauchy-Riemann equations for $f$ at $z$, the proof is complete.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 14:12

























answered Nov 29 '18 at 21:41









DidDid

247k23222457




247k23222457








  • 1




    $begingroup$
    (+1) Clear and concise. Well written.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:41














  • 1




    $begingroup$
    (+1) Clear and concise. Well written.
    $endgroup$
    – Mark Viola
    Nov 30 '18 at 4:41








1




1




$begingroup$
(+1) Clear and concise. Well written.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:41




$begingroup$
(+1) Clear and concise. Well written.
$endgroup$
– Mark Viola
Nov 30 '18 at 4:41


















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