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Assume that $ f: \mathbb R \to \mathbb R $ is uniformly continuous. prove that there are constants $A,B$ such that $ |f(x)| \le A + B|x| $ for all $ x \in \mathbb R $.

my concern is just $f$ is uniformly continuous on $\mathbb R$ how this is going to help us to find such $A$ and $B$.

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  • $\begingroup$ try to prove by contradiction. Suppose that there is no such $A$ and $B$. and use the definition of uniform continuity to obtain the contradiction. $\endgroup$ – Jane Jan 2 '16 at 3:32
  • $\begingroup$ how would we get a contradiction? could you please explain for me @ Jane $\endgroup$ – user145993 Jan 2 '16 at 3:38
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Since $f$ is uniformly continuous, there exists $c>0$ such that $\mid x-y\mid <c $ implies that $\mid f(x)-f(y)\mid <1$.

Let $x\in R$, write $x=nc+r, n$ is an integer, $0<r<c$, We have $\mid f(x)-f(0)\mid =\mid f(nc+r)-f(0)\mid= \mid f(nc+r)-f((n-1)c+r)+f((n-1)c+r)-...+f(r)-f(0)\mid\leq\mid f(nc+r)-f((n-1)c+r)\mid+....+\mid f(r)-f(0)\mid\leq \mid n\mid+1<(1/c)\mid nc+r\mid+r/c+1=1/c\mid x\mid +r/c+1$

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Uniformly continuous implies $|f(x)-f(y)|<1$ for all $|x-y|<\delta$, for some $\delta>0$. This means in turn that, for any positive integer $n$,

$$|f(n\delta/2)-f(0)|\\=|f(n\delta/2)-f((n-1)\delta/2)+f((n-1)\delta/2)-f((n-2)\delta/2)+\cdots+f(\delta/2)-f(0)|\leq \\|f(n\delta/2)-f((n-1)\delta/2)|+|f((n-1)\delta/2)-f((n-2)\delta/2)|+\cdots+|f(\delta/2)-f(0)| \leq n$$ Furthermore, if $x\in [(n-1)\delta/2, n\delta/2]$, we also have $|f(x)-f(0)|<n$ (same reasoning but replacing the first difference by $f(x)-f((n-1)\delta/2)$). We have $|x|\geq (n-1)\delta/2$ in the interval indicated, and thus $n-1\leq (2/\delta)|x|$.

Every positive $x$ will be in some interval $[(n-1)\delta/2, n\delta/2]$. We thus get $|f(x)-f(0)|\leq B(|x|+\delta/2)$ for positive $x$, where $B=2/\delta$, and so $|f(x)|\leq B|x|+B\delta/2+|f(0)|$. Similar reasoning can handle $x<0$.

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If you put $x=0$ then you see that $A$ has to be (at least) the absolute value of $f(0)$. So after a shift you can just assume $f(0) = A = 0$. Then it's immediately obtained by contradiction.

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WLOG, we may suppose that $f(0) = 0$. Then I clam that $|f(x)| \le 1 + B|x|$.

By uniform continuity, there is $\delta > 0 $ such that for $|x - y| \le \delta$, $|f(x) - f(y) | \le 1$. Now take $B = \frac{1}{\delta}$. For any $x > 0$, we can piece together little intervals of size $\delta$ and by the above, the difference of $f$ on each interval is no more than $1$.

Doing it precisely, for $x > 0$, we take $k_x = [\frac{x}{\delta}]$, the greatest integer not exceeding $\frac{x}{\delta}$. Then

$$|f(x)| = |f(x) - f(0)| \le |f(x) - f(k_x \delta)| + \sum_{i=1}^{k_x} \left|f(i) - f((i-1) \delta)\right| \le 1 + k_x \le 1 + Bx$$

Symmetric arguments hold for $x < 0$ and so the result follows.

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  • $\begingroup$ Why is there no loss of generality for $f(0)=0$ ? $\endgroup$ – Learnmore Jan 2 '16 at 4:23
  • $\begingroup$ Apply the same reasoning to the function $f(x) - f(0)$. $\endgroup$ – Future Jan 2 '16 at 4:26

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