Suppose cp is a C1 function on R such that cp(x)+a and cp’(x)~b as xjoo. Prove or give a counterexample: b must be zero.

To address the question, we need to clarify the assumptions and notation:

• $\phi (x)$ is a C1C^1C1 function on $\mathbb{R}$.
• As $x\to \infty$, $\phi (x)+a$ converges to some limit (presumably, this implies $\phi (x)\to -a$).
• φ′(x)→b\varphi'(x) \to bφ′(x)→b as $x\to \infty$.

We need to determine whether $b$ must be zero.

Analyzing the Conditions

Given:

1. $\phi (x)$ is a C1C^1C1 function, so φ′(x)\varphi'(x)φ′(x) exists and is continuous.
2. $\phi (x)+a$ converges as $x\to \infty$, which we interpret as $\phi (x)$ approaching a constant value, specifically $-a$.
3. φ′(x)→b\varphi'(x) \to bφ′(x)→b as $x\to \infty$.

Let’s examine the implications of these statements.

Asymptotic Behavior of $\phi (x)$

If $\phi (x)+a\to L$ as $x\to \infty$ for some constant $L$, then $\phi (x)\to -a+L$. For simplicity, we can redefine $L$ so that $\phi (x)\to L$ as $x\to \infty$.

Now, consider the derivative φ′(x)\varphi'(x)φ′(x). If $\phi (x)$ approaches a constant $L$, the derivative φ′(x)\varphi'(x)φ′(x) must approach zero. This follows because if $\phi (x)$ were not approaching a constant, then $\phi (x)$ would not settle at a particular value, violating the assumption of convergence.

Formal Proof

To make this rigorous, we can use the definition of a limit:

1. Since $\phi (x)\to L$ as $x\to \infty$, for any $ϵ>0$, there exists an $M>0$ such that for all $x>M$, $|\phi (x)-L|<ϵ$.
2. φ′(x)→b\varphi'(x) \to bφ′(x)→b as $x\to \infty$ implies for any $\delta >0$, there exists $N>0$ such that for all $x>N$, ∣φ′(x)−b∣<δ| \varphi'(x) – b | < \delta∣φ′(x)−b∣<δ.

Now, if b≠0b \neq 0b=0, then φ′(x)\varphi'(x)φ′(x) is bounded away from zero for large $x$. This means $\phi (x)$ would be increasing or decreasing without bound, contradicting the fact that $\phi (x)$ approaches the constant $L$.

Therefore, $b$ must be zero.

Conclusion

Under the given conditions, $\phi (x)$ converging to a constant $L$ and φ′(x)\varphi'(x)φ′(x) having a limit as $x\to \infty$ lead us to conclude that the limit of φ′(x)\varphi'(x)φ′(x) must be zero. Thus, $b$ must be zero.