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Study the uniform convergence of $f_n(x) = \frac{1}{n^2 x^2} , x \in (0,1].$

I'm having troubles manipulating $n$ and $x$ when trying to show whether a sequence is uniformly convergent or not. This is what i' ve thought for showing that the sequence is not uniformly convergent:

$f_n$ converges pointwise to $0$.

Let $\epsilon_0=0.5$ and $n\in \mathbb{N}$. If we take $x=\frac{1}{n} \in (0,1]$, then $|f_n(x) - 0 |=|\frac{1}{n^2 \frac{1}{n}^2} |=1 > \epsilon_0.$

However, I don' t know if I can fix $n$ in order to find an x such that the inequality doesn't hold as n can be arbitrarily large. Is this correct?

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Your approach is correct: since $(\forall n\in\mathbb{N}):f_n\left(\frac1n\right)=1$, it follows from the definition of uniform convergence that $(f_n)_{n\in\mathbb{N}}$ does not converge uniformly to the null function. And since it converges pointwise to that function, it cannot converge uniformly to some other function.

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  • $\begingroup$ You were correct. Thanks for the answer. $\endgroup$ – Yagger Apr 15 '18 at 15:15
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Apr 15 '18 at 15:17

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