My teacher wrote the following on the board:

$\frac{d}{dx}: C^1(a,b)\rightarrow C^0(a,b)$

I thought that a 1st order differential operator takes a function which is continuous and maps it to some function space where we don't know the properties (i.e. I don't know that the derivative is continuous). To me, what he wrote looks like it takes a function with a continuous derivative and maps it to a continuous function. Would someone explain this to me more clearly?


1 Answer 1


$C^1(a,b)$ is the space of functions which have a continuous derivative on the interval $(a,b)$. This means that if $f(x)$ is a function in $C^1(a,b)$, then not only does the derivative, $f'(x)$, exist for all $x\in(a,b)$, but that derivative is continuous, i.e. $f'(x)$ is a continuous function. Another way to write this is as your teacher did: applying the derivative operator to an element of $C^1(a,b)$ by definition yields a continuous function, also known as an element of $C^0(a,b)$. Or, in symbols: $$\frac{d}{dx}:C^1(a,b)\to C^0(a,b)$$

  • $\begingroup$ Oh, what you're saying is that what he wrote was only concerning the operator acting on the C^1 space? $\endgroup$
    – user269711
    Apr 18, 2016 at 18:54
  • $\begingroup$ @user269711 Yes, he appears to be restricting it to just functions from $C^1(a,b)$. $\endgroup$ Apr 18, 2016 at 18:56
  • $\begingroup$ Oh, well then that makes perfect sense. I guess that I was just reading what he wrote oddly. Thanks! $\endgroup$
    – user269711
    Apr 18, 2016 at 18:57

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