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I've been thinking this problem for so long it doesn't make sense to me now. I've been checking the other questions asked here on MSE, about it, (like these: 1, 2, 3, 4).

I have to show where this sequence converges uniformly: $$f_k(x):=\frac {x^3+kx}{kx^2+1}$$ with $f_k:[0,\infty)\to \Bbb R$
So far I've found the pointwise convergence, and is $f(x)=\frac 1x$.

If we look at the plot, and also by the fact that its limit is $1/x$, is clear that if $x\in [-1,1]$ the function goes crazy high as $k\to \infty$, so maybe the sequence is not uniformly convergent there, but for $|x|>1$ it approaches zero, so my gut tells me that is uniformly convergent there, I don't know if I'm interpreting the plot correctly, my gut hurts :( .
Now if I want to apply the techniques of the other questions, the difficulties that I find are that $f$ is not a constant nor a function that depends only of $k$ and it doesn't have a global max, so I'm stuck on how to procede.

Also I've noticed that $f_k(\frac 1k)\to 1$ and $f_k(k)\to \infty$, when $k\to\infty$, but this confuses me, would it mean that if $\alpha>0$, then $f_k$ doesn't converge in $[0,\alpha]$? if so, what does this $f_k(k)\to \infty$ means?

Here's the plot of some points: $k=1, k=2, k=4, k=20, k=100$ plot

Community
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Ana Galois
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2 Answers2

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Careful. The $f_k$ converge pointwise to $x^{-1}$ if $x\neq 0$, but to $0$ if $x=0$. This tells you immediately that you'll have problems near the origin! On the other hand, your $f_k$ are unbounded on $[a,\infty)$ and $(-\infty,a]$ for any $a$. Can you show that on any bounded interval that is bounded away from the origin convergence is uniform? Can you show it cannot be uniform on any interval containing the origin or any unbounded interval?

Pedro
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Note that uniform convergence preserves continuity.

Pedro
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Mhenni Benghorbal
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