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I was reviewing my text from Real Analysis, and something occured to me that hadn't before, nor is it mentioned in the text. The way you usually show that a sequence $\{f_n\}$ of functions does not converges uniformly is to first find a candidate function $f$ to which $\{f_n\}$ converges pointwise. Then, since uniform convergence implies pointwise convergence, if the the function does not converge uniformly to $f$, it cannot converge uniformly to any $f$. My question is regarding uniqueness of the limits.

Uniform convergence is equivalent to convergence in $(X,\lVert\cdot\rVert_{\infty})$, so there is no question of uniqueness of limits in metric space.

But I'm not sure about pointwise convergence. Could it be, that other than $f$ there was another test function $g$ that $f_n$ converged to pointwise, but neglected to check that $f_n\to g$ uniformly?

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    $\begingroup$ Can a sequence of real numbers have more than one limit? If $f_n$ converges to pointwise to some function $f$, then $f$ is unique. $\endgroup$ – angryavian Jun 4 '16 at 18:16
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No, it is not possible, because limits are unique in $(\mathbb{R}, \lvert\cdot\rvert)$ as well. Suppose $(f_n)_n$ converges pointwise to both $f$ and $g$, and $f\neq g$.

Then there exists $x_0$ such that $f(x_0)\neq g(x_0)$. But by assumption (pointwise convergence), the sequence $(f_n(x_0))_n$ converges to both $f(x_0)$ and $g(x_0)$, and therefore... $f(x_0)=g(x_0)$, contradiction.

In short: when it exists, the pointwise limit is unique.

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