# Show differential operator is not bounded using definition of bounded operators

Let $$T:C^{1}_{[a,b]} \rightarrow C^{0}_{[a,b]}$$ with $$a be the differential operator defined as $$Tx=x’$$.

The practice exercise asks for the kernel and range of such operator and also a demonstration using the definition of bounded operators.

An operator on normed spaces is said to be bounded when $$\|Tx\| \leq c\|x\|$$ where $$c$$ is a real number.

I have already been through examples of continuous functions which show that it is not bounded. However, this one specifically asks to use the definition.

Your contribution would be much appreciated. Feel free to shut this down if you indicate a solution that is available out there.

• You should clarify the norm on $C^1.$ This is untrue if you give it the topology generated by the norm $||f||_{C^1}=||f||_\infty+||f'||_\infty$. Presumably, you're equipping $C^1$ with the infinity norm. – cmk May 26 at 14:16
• I am sorry, @cmk. I made a mistake, I’ll edit it now. Thanks. – upStoneLock May 26 at 14:33
• Since the norm is not defined in the practice exercise I suppose we can define one.@cmk – upStoneLock May 26 at 14:38
• Typically, it'll be unbounded if we give it the sup norm, as shown in my answer. Hope that helps. – cmk May 26 at 14:47
• @cmk, thanks for your help. Can I add that you take $f_n$ without loss of generality? If so, what argument supports it is actually general for all continuously 1-differentiable functions? I am very thankful! – upStoneLock May 26 at 14:52

Take $$[a,b]=[0,1],$$ let $$n\in\mathbb{N}$$ be arbitrary. Next, consider the sequence $$f_n(t)=t^n.$$ Then, $$Tf_n(t)=nt^{n-1}.$$ If we equip $$C^1$$ with the infinity norm, then $$||Tf_n||_{\infty}=n,$$ but $$||f_n||_\infty=1.$$ So, $$\frac{||Tf_n||_{\infty}}{||f_n||_\infty}=n.$$ By assumption, this is true for any $$n$$. Can you see why this means there is no $$c$$ for which $$||Tf_n||_\infty\leq c||f_n||_\infty?$$

Consider a sequence of functions: $$f_{\omega}(x) = \sin(\omega x)$$ where $$\omega \to \infty$$. The maximum norm of $$f$$ is: $$||f_{\omega}||_{\infty} = 1$$

However, the first derivative is: $$Tf_{\omega}(x)=\omega \cos(\omega x)$$ And the maximum norm of $$Tf$$ is: $$||TF_{\omega}||_{\infty} = \omega \to \infty$$

• Great, one, @Hume2. However, do you believe it’s possible to argument that without using an example function? – upStoneLock May 26 at 14:53

Let $$f_0$$ be a function such as $$f_0(a)=f_0(b)$$ and $$f_0'(a)=f_0'(b)$$. Let $$f_{n+1}$$ be the function $$f_{n}$$ repeated twice and squezed into the interval $$[a,b]$$.

Then $$||f_0||_{\infty}=||f_1||_{\infty}=||f_2||_{\infty}=...=||f_n||_{\infty}$$

but: $$||Tf_n||_{\infty}=2^n||Tf_0||_{\infty}$$

• thank you very much. It does make for a nice argument, I just have take it all in. Thanks a bunch, really. – upStoneLock May 26 at 15:42