Let A f = a f''' for $f\in C_0^3(\mathbb R)$ where $a$ - some constant. Is it possible to find $a$ such that $ \|\lambda f-A f\|\geq \|\lambda f\| $ for all $f\in C_0^3(\mathbb R)$ and all $\lambda>0$? Here is norm is the uniform on $\mathbb R$.
The inequality with a differential operator
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0@user1551: I guess, you're right. – 2011-09-30
2 Answers
I suppose you are looking for a nontrivial $a$. Divide both sides of your inequality by $\lambda$. So you are essentially asking if one of the followings is true:
(a) \|f-kf'''\|\ge\|f\| for all $f\in C_0^3(\mathbb R)$ and all $k>0$;
(b) \|f+kf'''\|\ge\|f\| for all $f\in C_0^3(\mathbb R)$ and all $k>0$.
The two statements are actually equivalent because for any $g\in C_0^3(\mathbb R)$, if we define $f(x)=g(-x)$, then $\|f\|=\|g\|$ and \|g+kg'''\|=\|f-kf'''\|. Combining the two, we see that your requirement is equivalent to
(c) \|f+kf'''\|\ge\|f\| for all $f\in C_0^3(\mathbb R)$ and all $k\in\mathbb{R}$.
It should be easy to construct a counterexample to (c). For instance, consider $f(x)=(x+1)e^{-x^2}$ and some small positive $k$. However, I cannot think of a beautiful example that is easy to verify.
Yes, choose $a = 0$. If you want $a\ne 0$ the answer is negative. For $\lambda = 0$ the inequality becomes $ Af = 0 $ that is true only if f''' = 0.
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0Thank you, I made a typo in the sign of the inequality. – 2011-09-30