Find the function to which this given sequence of functions converges

Question

Given $$f_n(x)=\begin{cases} n^2x& \text{ if }\; 0\leq x \leq \frac{1}{n} \\ -n^2x+2n& \text{ if }\;\; \frac{1}{n} < x <\frac{2}{n} \\ 0& \text{ if }\;\;\frac{2}{n}\leq x \leq 1 \end{cases}$$ Now the question here is i have to find the function where this sequence of functions converges ,But am facing problem with these intervals ,it is easy to find limits at end points $i.e$ at $0$ and $1$ ,but to deal with these fractional $1/n$ parts ,please give me a hint so that i can proceed further

Thankyou

Answer 1

See $$\int_{0}^{\infty} f_n(x) dx=1$$ for all $n \in \mathbb N$.

And , the function is like a triangle which has maximum at $\frac{1}{n}$ and this maximum value increases with $n$.

When $n \to \infty$, $f_n(x)$ becomes infinitely narrow and infinitely high . But the area enclosed by the function is always unity.

So, $\lim \limits_{n \to \infty} f_n(x) =\delta(x)$ for $x \in [0,1]$. Where $\delta(x)$ is the Dirac Delta Function.

Evaluating the integral:

$$\int_{0}^{\infty} f_n(x) dx$$

$$=\int_{0}^{\frac{1}{n}} n^2x dx+\int_{\frac{1}{n}}^{\frac{2}{n}} (-n^2x+2n) dx+\int_{\frac{2}{n}}^{1} 0 dx$$

$$= \frac{1}{2}+\frac{1}{2}+0=1$$

Answer 2

The intervals get smaller as $n$ increases. So if you choose $x>0$ and then let $n\to \infty$, eventually $x$ will be left outside the given intervals, meaning $f_n(x)$ becomes zero and remains zero for all larger $n$. In the case where $x=0$, $f_n(x)$ is always zero.

Thus $f_n(x) \to 0$ for all $x \geq 0$. However it is clear also that the maximum value of each $f_n$ is $n$ and therefore increases without limit, so the convergence is not uniform.

To visualise, picture the intervals shrinking and squashing up closer and closer to zero as the functions $f_n$ become sharper and higher spikes.

Convergence to the Dirac function is true, but only in a different space, that of generalised functions in which a particular and different definition of convergence applies. From the question I suspect it is not the intent to introduce it here.