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I saw this question in another post and I proved it differently than the others who answered. I was wondering if my proof works.

Problem

Let $f: \mathbb{R} \to \mathbb{R}$ be a function. Suppose that $f$ is differentiable, that $f(0)=1$, and that $|f'(x)| \leq 1$ for all $x \in \mathbb{R}$. Prove that $|f(x)| \leq |x|+1$ for all $x \in \mathbb{R}$.

My Proof

Assume the conclusion didn't follow from the hypotheses, namely that $\exists x\in \mathbb{R}$ such that $|f(x)|>|x|+1$. We can show this for $x=0$. Then we get $|f(0)|>1 \Longrightarrow 1>1$, contradiction.


And here's where I found the question Practice problem from Mean Value Theorem in Real Analysis

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  • $\begingroup$ What do you mean by "we can show this for $x=0$"? Also, where did you use that $f$ is differentiable with $|f'|\leq 1$? $\endgroup$ – PhoemueX Jul 22 '18 at 5:34
  • $\begingroup$ Well I'm proceeding by contradiction, so I'm assuming the conclusion fails in some case, namely the case where $x=0$. And I didn't see the need to use those conditions since the contradiction followed immediately, if I did the proof correctly $\endgroup$ – john fowles Jul 22 '18 at 5:36
  • $\begingroup$ If you proceed by contradiction, you just get that the conclusion fails for some $x$. You cannot just assume that $x=0$. Also, since the claim is false without the assumption on the derivative, your proof is necessarily wrong if you nowhere use that assumption. $\endgroup$ – PhoemueX Jul 22 '18 at 5:38
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How can you say that $x=0$ is the point where $|f(0)|>|0|+1$?

We assume that there exists a point, $x_0$ such that $|f(x_0)|>|x_0|+1$, then we look at $$\frac{f(x_0)-f(0)}{x_0-0}$$ and you use this to prove that there is a point where $|f'(x)|>1$.

You can't just say "$x_0=0$", because the negative of all is exists, so it may be $0$, but it is also may be $10,5.4,777,\pi$ and so on.

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  • $\begingroup$ So when proving a contradiction you negate the conclusion and show that the negated conclusion, with the given hypothesis, lead to a contradiction. But I began by negating the conclusion, $|f(x)|>|x|+1$, then I used the conditions to find a contradiction, namely the condition that $f(0)=1$. $\endgroup$ – john fowles Jul 22 '18 at 6:24
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    $\begingroup$ @johnfowles what you did is assuming that a specific value of $x_0$ satisfy the negative of $|f(x)| \leq |x|+1$, but you need to show that if a value, not specific one, some arbitrary $x_0$, satisfy the negative of $|f(x)| \leq |x|+1$ then you have contradiction $\endgroup$ – Holo Jul 22 '18 at 6:35
  • $\begingroup$ ok, this makes sense $\endgroup$ – john fowles Jul 22 '18 at 6:39

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