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$C^{1}$ is the space of continuously differentiable functions and $C^{0}$ the space of continuous functions, both defined on the real interval $[0,1]$. Both spaces equipped with $d(f,g) = \sup_{x \in [0,1]|f(x) - g(x)|}$ as metric. $f^{1}(x)$ is the first derivative of f.

Now, is this operator uniformly continuous? It seems so but going about this demonstration became too confusing for me. Could you guys give me any hints, please?

nandevers
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  • Look at the zero function and a function with graph being zero everywhere except for a little, but very narrow, peak. – logarithm May 06 '19 at 20:00
  • Are you sure it's continuous? Since it's linear, there is a fairly easy way to check continuity... – ryan221b May 06 '19 at 20:01
  • Indeed, @ryan221b, I saw some examples like $f_{n} = sin(nx)$ which totally contradict my intuition about continuity. I'd like to develop a demonstration without using this. – nandevers May 06 '19 at 20:13
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    What are the norms on the two spaces? If $C^1$ is considered as a subspace of $C^0$ with the induced norm, see the above comments. However, if $C^1$ has the norm $||f|| = \sup_{[0,1]}|f(x)| + \sup_{[0,1]}|f'(x)|$, then it is simple to show that $T$ is uniformly continuous. – Jane Doé May 06 '19 at 20:13
  • @JaneDoé, that's the sup metric indeed, I'll add it to the question. – nandevers May 06 '19 at 20:16
  • OK, then existence of functions that are "close to 0" but with a "large derivative at some point" would show that $T$ is not continuous, wouldn't it? Do such functions exist? – Jane Doé May 06 '19 at 20:20
  • I found this example of a sequence $f_{n} = \sin(nx)$ which would fall in that category. I'll try and write a formal proof without using it as counterexample. Any hints in that direction? – nandevers May 06 '19 at 20:28

3 Answers3

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The continuity of the linear map

$f \mapsto f': C^1[0, 1] \to C[0, 1] \tag 1$

depends on the norms placed on $C^1[0, 1]$ and $C[0, 1]$. If, as is typical, we define

$\Vert \cdot \Vert _1: C^1[0, 1] \to \Bbb R \tag 2$

via

$\Vert f \Vert_1 = \displaystyle \sup_{x \in [0, 1]} \vert f(x) \vert + \sup_{x \in [0, 1]} \vert f'(x) \vert, \tag 3$

and

$\Vert \cdot \Vert_0: C[0, 1] \to \Bbb R \tag 4$

by

$\Vert f \Vert_0 = \displaystyle \sup_{x \in [0, 1]} \vert f(x) \vert, \tag 5$

then the mapping (1) is in fact continuous, since it is a bounded linear map:

$\Vert f' \Vert_0 = \displaystyle \sup_{x \in [0, 1]} \vert f'(x) \vert \le \displaystyle \sup_{x \in [0w, 1]} \vert f(x) \vert + \sup_{x \in [0, 1]} \vert f'(x) \vert = \Vert f \Vert_1. \tag 6$

However, if we use the metric

$d(f, g) = \displaystyle \sup_{x \in [0, 1]} \vert f(x) - g(x) \vert \tag 7$

which is in fact the $\Vert \cdot \Vert_0$ norm,

$d(f, g) = \Vert f - g \Vert_0, \tag 8$

on both $C[0, 1]$ and $C^1[0, 1]$, then $f \to f'$ is not continuous, since a function may be $\Vert \cdot \Vert$-small yet have an arbitrarily large derivative, for example

$f_k(x) = a\sin (kx) \tag 9$

is bounded in absolute value by $\vert a \vert$, but

$\vert f_k'(x) \vert = \vert ak\cos(kx) \vert \tag{10}$

becomes arbitrarily large as $k \to \infty$.

Robert Lewis
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This function is not continuous. For you can take any function on the real line and rescale the parameter ($f(x) \mapsto f(rx)$ for large $r$) to make the derivative large without affecting the $C^0$ norm of the function.

For a specific example, consider $f_n(x) = n^{-1} \sin (n^2x)$. As $n$ gets large, $f_n(x)$ goes to $0$ in $C^1(0)$ with the uniform norm, but $\frac{d}{dx}f_n(x) = n \cos(n^2x)$ diverges.

hedgehog enthusiast
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I think this might be a bit long for a comment, so I post as an answer.

In reply to your intuition about continuity, linearity “simplifies” continuity in that you only have to check continuity at 0 to determine continuity on the whole (vector) space. This is because

$$\|T(f) - T(g)\| = \|T(f-g)\| = \|T(h)\|$$

since we can replace $f-g$ by any other vector $h$.

From that it's easy to check that continuity is equivalent to boundedness: there is a constant $M > 0$ such that

$$ \|T(f)\| \leq M\| f \|$$

for all $f$.

So to show that a (linear) operator between normed spaces is continuous, all you have to do is estimated a suitable $M$. On the other hand, to show discontinuity it is enough to pick a sequence of functions $(f_n)_n$ that violate the inequality, say

$$ \| T(f_n) \| \geq n \| f_n \| $$

(You need infinitely many: if you only had a finite number, then you could pick the biggest norm and you'd be bounded again.)

In general, derivative operators are not bounded on $C^1([0,1])$: just think of polynomials $(x^n)_n$.

ryan221b
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