show Lebesgue dominated convergence theorem fails for ${n^2xe^{-nx}} x\in [0,1]$

Question

show Lebesgue dominated convergence theorem fails for the sequence of functions $f_n=n^2xe^{-nx}$ $x\in [0,1]$

Here is my solution. Is it correct?

• $f_n$ is an integrable function

• the sequence converges pointwise to zero

• Now we try to find an integrable function $g(t)$ such that $|f_n(t)|\leq g(t)$ First, at each n, we find the point x where $f_n$ attains its maximum. $$f'_n(x) = n^2e^{-nx}-n^3xe^{-nx}=0 \implies x = 1/n$$ then, $max(f_n) = f_n(1/n) = ne^{-1}$

since, the maximum point increases whenever n increase, we can't find a single $g(t)$ that will bound all $f_n$. Hence, the theorem fails for this sequence.

Answer 1

The first two points are correct, and you can use an approach as yours to show $f_n$ cannot be dominated; but there is another (simpler) way than this third step. To show the theorem fails, indeed, it is sufficient to show its conclusion does not hold (this by contradiction will in particular imply that the function cannot be dominated, otherwise the theorem and its conclusion would apply).

Specifically, it suffices to show that $\int_{[0,1]} f_n$ does not converge to $0$, as $\int_{[0,1]} \left( \lim_{n\to\infty} f_n(x)\right) dx = \int_{[0,1]} 0 = 0$. But this follows by observing that $$ \int_{[0,1]} f_n = \int_{[0,1]} n^2x e^{-nx} dx = 1-(n+1)e^{-n} \xrightarrow[n\to\infty]{} 1 $$