Suppose $f(x)=x^i$ as a simple polynomial function of power $i$. As we know from the definition of the derivative operators, such an operator must be linear. How can we prove that the derivative operator of polynomials is actually a linear operator?
I am specifically interested in using Fréchet derivative formula to give a proof.

  • $\begingroup$ You can actually show that the derivative is linear for every function using the limit definition. $\endgroup$ – Javi Aug 11 '18 at 9:43
  • $\begingroup$ @Javi but the limit definition of derivative requires the assumption of the linearity of the operator; it can not be used to show linearity. $\endgroup$ – Mehdi Moghadasian Aug 11 '18 at 10:01
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    $\begingroup$ It only uses linearity of the limit, which can also prove if you dare. $\endgroup$ – Javi Aug 11 '18 at 10:10
  • $\begingroup$ @Javi Thanks for your response. Consider Fréchet derivative. If there exists a bounded linear operator $D(.): V\rightarrow U$ such that $\lim_{h\to 0}\frac{||f(x+h)-f(x)-Df(x)h||}{||h||}=0$ holds, then $D(.)$ is the unique derivative of $f(x)$. Do you mean if this operator exists, it is necessarily a linear operator? Or you are suggesting to show the linearity by using the definition? If latter is what you mean, how can I show the linearity principle $D(rf+sg)(x)=rDf(x)+sDg(x)$ by using the limit definition? Many thanks for your help. $\endgroup$ – Mehdi Moghadasian Aug 11 '18 at 10:56
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    $\begingroup$ Many thanks @Javi . Yes, I'd forgotten to mention Frechet. I edited my question. $\endgroup$ – Mehdi Moghadasian Aug 11 '18 at 14:35

\begin{align} &\lim_{h \to 0} \frac{|r(x+h)^i+s(x+h)^j-(rx^i+sx^j)-(irx^{i-1}-jsx^{j-1})h|}{|h|}\\ &=\lim_{h \to 0}\left|r\left(\frac{(x+h)^i-x^i}{h} -ix^{i-1}\right)+s\left(\frac{(x+h)^j-x^j}{h} -jx^{j-1}\right)\right|\\ &=\left|r\lim_{h \to 0}\left(\frac{(x+h)^i-x^i}{h} -ix^{i-1}\right)+s\lim_{h \to 0}\left(\frac{(x+h)^j-x^j}{h} -jx^{j-1}\right)\right|\\ &=0 \end{align}

where I have used continuity of norm and linearity of limit.


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