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Let $f\colon (0,1]\to [-1,1]$ be a continuous function. Let us define a function $h$ by $h(x)=xf(x)$ for all $x$ belongs to $(0,1]$. Prove that $h$ is uniformly continuous.

We know $f$ is uniformly continuous on $I$ if $f'(x)$ is bounded on $I$. Here $h'(x)= xf'(x) + f(x)$ and $f(x)$ is bounded here. How can I prove that $xf'(x)$ is bounded here. Please help me to solve this. Thanks in advance.

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  • $\begingroup$ just using the eps/delta definition here is more easy $\endgroup$ – user392395 Apr 30 '17 at 15:39
  • $\begingroup$ It is better to use mathjax to format equations. $\endgroup$ – Arbuja Apr 30 '17 at 15:45
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$f $ continuous at $(0,1] \implies $

$$g:x\mapsto \frac {f (x)}{x} $$ continuous at $(0,1] $ but this doesn't mean that

$$x\mapsto xg (x)=f (x) $$ is uniformly continuous at $(0,1] $.

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Hints: 1. Since $f$ is only given to be continuous, thinking about derivatives of $f$ can't possibly be the way to go. 2. Alternate approach: It's enough to show $\lim_{x\to 0^+} h(x)$ exists, because then $h$ can be regarded as continuous on $[0,1].$

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Let $\epsilon > 0$ be given and $x_0$ be some point in the interval. First note $\max f(x) = 1$ and $\max x = 1$ given in the problem. Since $f$ is continuous, there exists a $\delta>0$ such that $|x-x_0|<\delta \implies |f(x) - f(x_0)| < \frac{\epsilon}{3}$. Choose $\delta=\frac{\epsilon}{3}$.

$$ \begin{align} \left|h(x) - h(x_0)\right| &= \left|xf(x) - x_0f(x_0)\right| \\ &= |xf(x) - xf(x_0) + xf(x_0) - x_0f(x_0)| \\ &\leq |x||f(x) - f(x_0)| + |f(x_0)||x-x_0| \\ &\leq |f(x) -f(x_0)| + |x-x_0| \\ &< \frac{\epsilon}{3} + \delta = \frac{\epsilon}{3} + \frac{\epsilon}{3} < \epsilon \end{align}$$

Since $\delta$ is only dependent on $\epsilon$ and independent of $x_0$, $h$ is therefore uniformly continuous.

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  • $\begingroup$ yeah..it is clear now. Thanks to all $\endgroup$ – SANDIP SHARMA Apr 30 '17 at 18:50

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