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Let $Ax(t)=x(\sqrt{t})$ in $C^1[0,1]$ with norm $\|x\|_1=\max\limits_{t\in[0,1]}|x(t)|+\max\limits_{t\in[0,1]}|x'(t)|$. It is required to check it for boundedness.

All I know is that the operator is not defined on the whole space, but is defined on the set of functions that satisfy the condition $\lim\limits_{t\to+0}\dfrac{x'(t)}{t}=A\in\mathbb{R}.$

I can't get an estimate like $\|Ax\|\leq c\|x\|$, but I also can't find the sequence $x_n(t)$ in unit ball, such that $\|Ax_n(t)\|\to\infty.$

I will be grateful for any help or hint on how to proceed.

thing
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1 Answers1

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I believe $A$ is unbounded. We have

$$\|Af\| = \sup_{x\in[0,1]} |f(\sqrt{x})| + \sup_{x\in[0,1]} \left|f'(\sqrt{x})\cdot \frac1{2\sqrt{x}}\right| = \sup_{x\in[0,1]} |f(x)| + \frac12\sup_{x\in[0,1]} \left|\frac{f'(x)}x\right|$$

and in general we cannot dominate the last expression by $\|f\|_\infty$ and $\|f'\|_\infty$, even if $\lim_{x\to0^+} \frac{f'(x)}x$ is finite.

For $n \ge 2$ define $f_n(x) = -\cos(n\pi x)$. Then $f_n \in C^1[0,1]$ and $$f_n'(x)=n\pi \sin(n\pi x) \implies \lim_{x\to 0^+} \frac{f_n'(x)}{x} = \lim_{x\to 0^+} \frac{n\pi \sin(n\pi x)}{x} = n^2\pi^2$$ so $f_n$ is admissible. We have $$\sup_{x\in[0,1]} |f_n(x)| = 1, \quad \sup_{x\in[0,1]} |f_n'(x)| = f_n'\left(\frac2{n\pi}\right) = n\pi, \quad \sup_{x\in[0,1]} \left|\frac{f_n'(x)}x\right| = n^2\pi^2$$

and hence $$\|f_n\| = 1+n\pi, \quad \|Af_n\| = 1+\frac12n^2\pi^2 \implies \frac{\|Af_n\|}{\|f_n\|} = \frac{1+\frac12n^2\pi^2}{1+n\pi} \xrightarrow{n\to\infty} +\infty.$$

mechanodroid
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