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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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Divide both sides 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 $k\ge0$; (b) $\|f+kf'''\|\ge\|f\|$ for all $f\in C_0^3(\mathbb R)$ and all $k\ge0$. –  user1551 Sep 30 '11 at 8:40
    
@user1551: I guess, you're right. –  Ilya Sep 30 '11 at 8:52

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up vote 3 down vote accepted

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. –  Ilya Sep 30 '11 at 8:07

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