Show the following statement

Question

for each $ n \geq1, $ let $ f_{n} (x) = \dfrac{nx}{n^{2}x^{4}+1}$. Consider the function $ f $ given by, $$ \lim_{n \rightarrow \infty} f_{n}(x) = f(x) $$ a) Verify that $$ \int_{0}^{1} \left(\lim_{n \rightarrow \infty} f_{n}(x) \right) dx \neq \lim_{n \rightarrow \infty} \int_{0}^{1} f_{n} (x) dx $$ b) is $ f_{n} $ uniformly convergent to $ f $ in $ [0,1] $? Justify.

I have some ideas, but I'm not sure ... At least for the second literal. The first literal, if I take the limit of $ f_{n} (x) $ gives me zero, and well, solving the integral is interesting. Help for literal b)

Answer 1

If $f_{n}\rightarrow f$ uniformly, then given $\epsilon>0$, there is an $N$ such that $|f_{n}(x)-f(x)|<\epsilon$ for all $x\in[0,1]$ and $n\geq N$, then \begin{align*} \left|\int_{0}^{1}f_{n}(x)dx-\int_{0}^{1}f(x)dx\right|\leq\int_{0}^{1}|f_{n}(x)-f(x)|dx\leq\int_{0}^{1}\epsilon=\epsilon, \end{align*} this shows that \begin{align*} \int_{0}^{1}f(x)dx=\lim_{n\rightarrow\infty}\int_{0}^{1}f_{n}(x)dx, \end{align*} which contradicts the first result.

Answer 2

You can also show directly that $f_n$ cannot converge uniformly to the pointwise limit $f(x) = 0$ on $[0,1]$. It is a bit cumbersome in comparison with the far more elegant argument using the integral:

• Calculating $f'_n(x) = \frac{n-3n^3x^4}{(n^2x^4+1)^2}$ you find, that $f_n$ has a lokal maximum at $x_n = \frac{1}{\sqrt[4]{3n^2}}$ with $f_n(x_n) = \frac{\sqrt[4]{3^3}}{4}\sqrt{n}> \frac{\sqrt{n}}{4}$.
• $\Rightarrow$ For $n\geq 16$ you have $f_n(x_n) > 1$ with $x_n \in (0,1)$.

Now, you can argue by contradiction. If we had $f_n \Rightarrow 0$, then for the specific $\epsilon_0 = 1$ we should find an $N_0$ such that for all $n \geq N_0$ and all $x \in [0,1]$ $$|f_n(x)-0| = f_n(x)< 1 \Rightarrow f_n(x_n) < 1 \mbox{ for all } n \geq N_0$$ $$\mbox{ Contradiction to } f_n(x_n) > 1 \mbox{ for all } n \geq 16$$ Done.