# For $f_n(x)=nx e^{-nx^2}$ can operations of integration and limit be interchanged?

For $$f_n(x)=nx e^{-nx^2}$$ can operations of integration and limit be interchanged?

$$\lim_{n \to \infty} f_n(x)\, dx = 0 \Rightarrow \\ \int_0^1\lim_{n \to \infty} f_n(x)\, dx = 0 \\ \\ \int_0^1 f_n(x)\, dx = \left.-\frac{1}{2} e^{-nx^2} \right|_{0}^{1} = -\frac{1}{2}(e^{-n}-1) \Rightarrow \\ \lim_{n \to \infty} \int_0^1 f_n(x)\, dx = \frac{1}{2} \\ \lim_{n \to \infty} \int_0^1 f_n(x)\, dx \ne \int_0^1\lim_{n \to \infty} f_n(x)\, dx$$ So, from calculating each side can conclude that the operations of integration and limit can not be interchanged.

But the following can lead to a different conclusion: \begin{align} \left| \int_0^1 f_n(x)\, dx - \int_0^1\lim_{n \to \infty} f_n(x)\, dx \right| &= \left| \int_0^1(f_n(x)- \lim_{n \to \infty} f_n(x))\,dx \right| \\ &<= \int_0^1 \left |f_n(x)- \lim_{n \to \infty} f_n(x) \right | \,dx \end{align} For any $$x\in (0,1)$$ $$f_n(x)$$ converge as $$n\to \infty$$. $$\lim_{n\to\infty} f_n(x) = f$$ So, for $$\forall \epsilon>0\,$$ $$\exists N > 0$$ for $$n \ge N$$

$$\left |f_n(x)- f \right | < \epsilon \Rightarrow \\ \int_0^1 \left |f_n(x)- \lim_{n \to \infty} f_n(x) \right | \,dx < \epsilon$$ Can get that for $$\forall \epsilon>0\,$$ $$\exists N > 0$$ for $$n \ge N$$ $$\left| \int_0^1 f_n(x)\, dx - \int_0^1\lim_{n \to \infty} f_n(x)\, dx \right| < \epsilon \Rightarrow \\ \lim_{n \to \infty} \int_0^1 f_n(x)\, dx = \int_0^1\lim_{n \to \infty} f_n(x)\, dx$$ which contradicts with the conclusion get before.

Let $$f_n(x)=nxe^{-nx^2}$$ and $$f(x)=\lim_{n\to \infty }f_n(x)=0$$.

So, it is true that for each $$x\in [0,1]$$, given any $$\varepsilon>0$$, there exists a number $$N(x,\varepsilon)>0$$ that can depend on both $$x$$ and $$\varepsilon$$, such that whenever $$n>N(x,\varepsilon)$$, $$|f_n(x)-f(x)|<\varepsilon$$. We call this pointwise convergence of $$f_n(x)$$.

But note that we cannot say that for each $$\varepsilon>0$$, there exists a number $$N(\varepsilon)$$ such that whenever $$n>N(\varepsilon)$$ and $$x\in [0,1]$$, $$|f_n(x)-f(x)|<\varepsilon$$. We call this type of convergence uniform convergence.

That $$f_n(x)$$ fails to converge uniformly can be seen by observing that

$$\sup_{x\in[0,1]}|f_n(x)|=\sup_{x\in[0,1]}|nxe^{-nx^2}|=\sqrt{\frac n2}e^{-1/2}$$

and $$\lim_{n\to\infty}\sup_{x\in[0,1]}|f_n(x)|= \infty$$.

Hence given any $$\varepsilon>0$$, we cannot find a number $$N(\varepsilon)$$ that is independent of $$x$$ such that $$|f_n(x)-f(x)|<\varepsilon$$. And we cannot write, therefore, that for all $$\varepsilon>0$$ there exists a number $$N(\varepsilon)$$ such that whenever $$n>N$$, $$\int_0^1 |f_n(x)-f(x)|\,dx<\varepsilon$$.