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 $[0,1]?$

  • 1
    $\begingroup$ $Y$ 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

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 $\|Df_n\|\not\to 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.