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I'm trying to study the pointwise and uniform convergence of the sequence of functions $k^x\sin(kx)$.

It's easy to see that it converges pointwise to $0$ for all $x \le 0$, while it does not converge for positive values of $x$. I'm having trouble, however, with the uniform convergence: the text says the sequence converges uniformly on all intervals of the form $(-\infty,a]$, where $a < 0$. I don't understand why it doesn't converge uniformly at $0$.

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  • $\begingroup$ Check the convergence along the sequence $x_k = - 1/ln(k)$ ... $\endgroup$
    – xounamoun
    Jun 18, 2023 at 15:41

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Define $f_k(x) = k^x \sin(kx)$. On the interval $I = (-\infty,0]$ consider the sequence $x_k = -1/k$. One can check that $f_k(x_k) \to\sin(-1)$. This means that $f_k$ can not converge uniformly to the zero-function on $I$.

Note that this construction would not work on $I = (-\infty, a)$ with $a<0$, as at some point $x_k$ would not lie in $I$.

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  • $\begingroup$ Thank you, this is helpful, though I'm not sure how I'd come up with this if I didn't know I was wrong. The way I learned it is I'm supposed to check $\sup|k^x\sin(kx)|$ and if it's $0$ then there is uniform convergence. I considered $|k^x\sin(kx)|\le|k^x|$, which is $1$ if $x=0$, but $\sin(k0)=0$; and $|k^x|=0$ as $n$ approaches infinity for all negative values of $x$. That's why I thought there was uniform convergence at $0$. $\endgroup$
    – VodkaTonic
    Jun 20, 2023 at 13:03

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