# Studying pointwise and uniform convergence of $k^x\sin(kx)$

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 $$0$$.

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

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 $$I$$.
• 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 $0$. Jun 20, 2023 at 13:03