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In my text there's a problem which reads as:

Consider $C[0, 1]$ with the norm $\|.\|_\infty$. Let $Y$ be the vector subspace of all differentiable functions on $[0, 1].$ Consider the linear map $D: (Y, \|.\|_\infty)\to(C[0, 1],\|.\|_\infty)$ given by $Df = f'$, the derivative of $f$. Show that $D$ is not continuous.

Is $D$ well-defined for what is the guarantee of the continuity of the derived function on $[0,1]$ of a continuous function on $[0,1]?$

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    $\begingroup$ $Y$ should probably have been defined as the subspace of all continuously differentiable functions on $[0,1]$. $\endgroup$ – Brian M. Scott Aug 19 '13 at 6:23
  • $\begingroup$ @BrianM.Scott: I think once $Y$ is defined for continuously differentiable functions the argument can be made as: Consider $\left(f_n=\dfrac{1}{n}x^n\right)_n\subset Y.$ Then $\|f_n-0\|_\infty=\|f_n\|_\infty=\sup\{|f(x)|:x\in[0,1]\}$$=\sup\left\{{\dfrac{1}{n}}x^n:x\in[0,1]\right\}=\dfrac{1}{n}\to0\implies f_n\to0$ in $Y.$ But $\|Df_n-D0\|_\infty=\|Df_n\|=\|x^{n-1}\|\to 1\ne0.$ $\endgroup$ – Sriti Mallick Aug 19 '13 at 6:59
  • $\begingroup$ The idea’s good, but the last sentence needs a little work: $\|Df_n\|=\sup_{x\in[0,1]}|x^{n-1}|=1$. $\endgroup$ – Brian M. Scott Aug 19 '13 at 7:03
  • $\begingroup$ @BrianM.Scott: Thanks so much. I could have edited that unless there was a time frame of 5 minutes. $\endgroup$ – Sriti Mallick Aug 19 '13 at 7:07
  • $\begingroup$ You’re welcome. $\endgroup$ – Brian M. Scott Aug 19 '13 at 7:08
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As Brian M. Scott said, the map $D:Y\to C[0,1]$ is defined provided we take $Y$ to be the space of continuously differentiable functions, usually denoted $C^1[0,1]$.

Concerning the discontinuity of $D$, a standard example is $f_n(x)=n^{-1}\sin nx$. You can check that $\|f_n\|\to 0$ but $\|Df_n\|\not\to 0$.

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