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Please help me check, if

  1. $f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n + g_n$ uniformly converge to $f+g$

  2. $f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n \cdot g_n$ uniformly converge to $f\cdot g$

What I've done:

I Guess that first statement is true, second is false. I've tried to calculate directly from definition, nothing seems correct.

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  • $\begingroup$ 1. is true, but 2. is false. $\endgroup$ – Julien Apr 3 '13 at 17:04
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    $\begingroup$ @SamiBenRomdhane: As I mentioned below, uniform convergence does not imply the functions are bounded. $\endgroup$ – copper.hat Apr 3 '13 at 17:23
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    $\begingroup$ @SamiBenRomdhane Uniform convergence is not restricted to bounded functions. For any choice of function $f(x)$, the functions $f_n(x)=f(x)+1/n$ converge uniformly to $f$. $\endgroup$ – Julien Apr 3 '13 at 17:29
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For the second case, take $f_n(x)=g_n(x) = x+\frac{1}{n}$. With $f(x)=g(x) = x$, we see that $f_n \to f, g_n \to g$ uniformly.

However, $f_n(x) g_n(x) = f(x)g(x) + \frac{2}{n} x + \frac{1}{n^2}$, hence the convergence is not uniform on unbounded sets.

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  • $\begingroup$ @user57: Thanks for catching that! $\endgroup$ – copper.hat Apr 3 '13 at 17:03
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    $\begingroup$ @SamiBenRomdhane: Yes, it is correct. Dominic Michaelis's deleted (why?) answer shows that if $f,g$ are bounded, then the convergence is uniform. If the functions are continuous, and the domain a closed interval, then this is satisfied. $\endgroup$ – copper.hat Apr 3 '13 at 17:18
  • $\begingroup$ What about if we have the countable sum of $f_n^{(1)} + f_n^{(2)} + \ldots$? Is this uniformly convergent? $\endgroup$ – jjjjjj Jun 3 '17 at 19:04
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The second is not corrent try $f_n(x)= \frac{1}{n}$ and $g_n(x)=x$ which both converge uniformly. Now their product is $h_n(x)= \frac{x}{n}$ which converge pointwise to the zero function. But $$\sup _{x \in \mathbb{R}}|h_n(x)-0|=\infty \forall n\in \mathbb{N}$$ Therefore it cannot converge uniformly to the zero function.

Notice that the only thing that was keeping you from applying the triangle inequality was the boundness of $f_n$ and $g_n$.

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    $\begingroup$ @SamiBenRomdhane The function may not be bounded but the convergence is uniformly because $$\sup _{x\in \mathbb{R}}|g_n(x)-x|=0$$ which porve the $g_n$ converges uniformly to $x$ on $\mathrm{R}$. $\endgroup$ – clark Apr 3 '13 at 17:20
  • $\begingroup$ @SamiBenRomdhane: Uniform convergence does not require bounded functions. $\endgroup$ – copper.hat Apr 3 '13 at 17:21
  • $\begingroup$ $f_n \to f$ uniformly iff $\sup_x |f_n(x)-f(x)| \to 0$. You can choose to restrict to bounded functions if you want, but it is not required. $\endgroup$ – copper.hat Apr 3 '13 at 17:29
  • $\begingroup$ Clark, how can $g_n(x) = x$ converge uniformly but $h_n(x) = x/n$ does not? $\endgroup$ – user104235 Nov 20 '13 at 2:28
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    $\begingroup$ @user104235 When I say converges uniformly here, as a sequence of function $g_1, g_2 , \cdots $ which of course will converge to the function $x$ because it is constant when viewed as a sequence of functions. But $h_n(x) = \frac{x}{n}$ converges pointwise to the zero function , again as a sequence of functions but there is too infinite mass escaping for each fixed $n_0$ $\endgroup$ – clark Nov 20 '13 at 5:49
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Let $\epsilon >0$ there's $n_1\in \mathbb{N}$, $\forall n\geq n_1$ $||f_n-f||_\infty<\epsilon/2$ and there's $n_2\in \mathbb{N}$, $\forall n\geq n_2$ $||g_n-g||_\infty<\epsilon/2$,

now let $n_0=\max(n_1,n_2)$ then $\forall n\geq n_0$ $||f_n+g_n-(f+g)||_\infty\leq ||f_n-f||_\infty+||g_n-g||_\infty <\epsilon$

For the second result if we suppose that the functions are continuous on the compact $[a,b]$ then use the same idea with the fact that $(f_n)$ is bounded since it's convergent sequence and with the inequality $$||f_n g_n - fg||_\infty=||f_n g_n - f_n g + f_n g -fg||_\infty\leq ||f_n||_\infty||g_n-g||_\infty + ||g||_\infty||f_n-f||_\infty$$

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    $\begingroup$ $f,g$ need not be bounded. $\endgroup$ – copper.hat Apr 3 '13 at 17:22

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