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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 $\mathbb R$.

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    @user1551: I guess, you're right.2011-09-30

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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.

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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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    Thank you, I made a typo in the sign of the inequality.2011-09-30