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Let $f$ be entire, $|f(z)|\leq 3|z|^4+1$ prove $f$ is a polynomial with deggre $\leq 4$

Similar questions have been asked here, but non answer satisfies me, so I tried a way I get, I just want someone to verify this.

$f^{(4)}(z_0)=\frac{4!}{2\pi i} \int_c\frac{f(z)}{(z-z_0)^5}dz$,

$|f^{(4)}(z_0)|\leq 4! \frac{3|z|^4+1}{|z|^5|(1-\frac{z_0}{|z|^5})|}|z|=4!\frac{3+\frac{1}{|z|^4}}{|(1-\frac{z_0}{|z|^5})|} $, where $|z|$ comes from the length of the line (instead of $r$)

leting $|z|\rightarrow \infty$ we get $|f^{(4)}(z_0)|\leq 72$ since $z_0$ is random from Liouville's theorem $f^{(4)}(z)$ is constant, it follows $f$ is a polynomial with deggre $\leq 4$

領域展開
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    In your first line, $z$ is the variable of integration. What is $z$ in your second line? – Jason May 18 '21 at 14:14
  • I am using the classic line integral inequality i am not sure what you mean [https://math.stackexchange.com/questions/1084584/inequality-for-line-integral] – 領域展開 May 18 '21 at 14:17
  • It's not clear what I am doing ? – 領域展開 May 18 '21 at 14:35
  • There is no integral in your second line. You have, essentially, stated the inequality $|\int g(x) dx| \le h(x)$. Do you see why that does not make sense? – Jason May 18 '21 at 16:37
  • This is correct i just did $|\int g(x)| \leq |g(x)|L $ – 領域展開 May 18 '21 at 17:13
  • That's precisely the problem. The left hand side of your inequality is a number - the $x$ variable is integrated out. The right hand side, on the other hand, is a function of $x$. I suggest you spend a little time becoming a little more familiar with integration inequalities, as at this point there are some clear gaps in your knowledge. – Jason May 18 '21 at 17:46
  • i dont know what you mean i just did exactly the same thing as the first answer here [https://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial?rq=1] with different symbols – 領域展開 May 19 '21 at 10:15
  • More precise the line integral is a real integral for example, $\left | \int_{g}^{}f \right |=\left | \int_{a}^{b} f(g(t))g'(t)dt\right |\leq \int_{a}^{b}\left | f(g(t))\left | g'(t) \right |dt \right |\leq (3|z|^{4}+1)\int_{a}^{b}\left | g'(t) dt\right |$ , unless this is wrong – 領域展開 May 19 '21 at 10:49
  • This is wrong, and I have explained why. Please read my responses more carefully. The left hand side of your inequality is a number - the result of integrating $f$. The right hand side is a function of $z$ - if you plug in different values of $z$, you will get different values for the right hand side. That does not make sense! You instead should have $|f(g(t))| \le 3|g(t)|^4+1$, inside the integral of course. Please spend some time on some simple, explicit examples, and then attempt your problem again. – Jason May 19 '21 at 14:23
  • Since $|f(z)|\leq k|z|^m$,

    $$|f^{(n)}(0)|\leq \frac{n!k|z|^m}{R^n}$$ for all $|z|=R$. For $n>m$, letting $R\rightarrow\infty$....., then how is this correct ?

     – 領域展開 May 19 '21 at 15:09
  • It's literally the same thing, instead of I write $|z-z_0|=R$ – 領域展開 May 19 '21 at 15:17

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