# Pointwise limits in Real Analysis 4

In the following two questions I need to find the pointwise limit of the sequence defined for $x\in R$. I have had some success with the first example and believe I have done it correct. But in the second example I am not sure how to find the pointwise limit or if it is uniformly convergent

1) $f_n(x) = \dfrac{nx+x^2}{n^2}$. Then $$\lim\limits_{n\to\infty} f_n(x) =\lim\limits_{n\to\infty} \left(\dfrac{x}{n} + \dfrac{x^2}{n^2}\right) = x\left(\lim\limits_{n\to\infty} \dfrac{1}{n}\right) + x^2\left(\lim\limits_{n\to\infty} \dfrac{1}{n^2}\right)=0+0=0$$

So this converge pointwise to $0$.

However how do I find the pointwise limit of the following sequence and whether or not that it uniformly converges?

2) $f_n(x) = e^{-(x-n)^2}$

• Observe that $$e^{-(x-n)^2}=e^{-(n-x)^2}=\frac1{e^{(n-x)^2}}$$ Hope it helps. – Masacroso Feb 8 '17 at 22:28
• Note that $f_n(n) = 1$ for all $n$. – copper.hat Feb 8 '17 at 22:33

For 1), uniform convergence would imply that given some $\epsilon > 0$, you can find $n$ large enough so that $|f_n(x)| < \epsilon$ for all $x$. Why is this impossible? (Consider large enough $x$.)

For 2), as $n \to \infty$ the exponent tends to $-\infty$ so the pointwise limit is $0$. For uniform convergence, follow the same approach as 1). For $0<\epsilon<1$, why can't you find $n$ large enough so that $|f_n(x)| < \epsilon$ for all $x$? (See copper.hat's hint: note that $|f_n(x)| \le |f_n(n)|=1$ for all $x$ and $n$.)

• could you further elaborate for the uniform convergence for number 2? trying to understand this logic – Bl409 Feb 8 '17 at 22:47
• what implicatations does $f_n = n$ for all n have? – Bl409 Feb 8 '17 at 22:55
• @Bl409 Suppose $\epsilon=1/2$ and you want to verify $|f_n(x)|<\epsilon$ for all $x$. This is impossible since $|f_n(n)|=1$. – angryavian Feb 8 '17 at 22:59
• so then this is not uniformly convergent since $|f_n(n)|=1$? – Bl409 Feb 8 '17 at 23:04

Try fixing $x_0\in \mathbb R$ and study $\lim_{n\to +\infty} f_n(x_0)$. In the case you are asking:

$$\lim_{n\to +\infty} e^{-(x_0-n)^2} = \lim_{n\to +\infty} \frac{1}{e^{(x_0-n)^2}}=0$$

because as $n\to +\infty$, as $x_0$ is fixed, $(x_0-n)^2$ becomes larger, and so does $e^{(x_0-n)^2}$.

If you do not see it clear, you can choose some particular values of $x_0$.

So $f_n$ converges pointwise to $0$. Now, to study uniform convergence, note that the only possible limit is $0$, because uniform convergence implies pointwise convergence. So you will have to prove that $$\sup_{x\in \mathbb R} |f_n(x)-0|=\sup_{x\in \mathbb R} |f_n(x)|\to 0$$ as $n\to \infty$.

But, in this case, $f_n(n)=1$, so $\sup_{x\in \mathbb R} |f_n(x)|>1$.

In the general case, it is usually a good idea to find $\sup |f_n(x)|$ using derivatives.

• so does thus prove that it is not uniformly convergent for number 2? – Bl409 Feb 8 '17 at 22:49
• @Bl409 Yes. If $\sup_{x\in \mathbb R} |f_n(x)| >1$ for all $n\in \mathbb N$, it cannot be $\sup_{x\in \mathbb R} |f_n(x)|\to 0$. – A. Salguero-Alarcón Feb 9 '17 at 15:13