# Let $f_n(x)=\frac{1}{nx+1}$ and $g_n(x)=\frac{x}{nx+1}$ for $x\in (0,1)$. Show that $f_n$ doesn't converge uniformly but that $g_n$ does

Let $$f_n(x)=\frac{1}{nx+1}$$ and $$g_n(x)=\frac{x}{nx+1}$$ for $$x\in (0,1)$$. Show that $$f_n$$ doesn't converge uniformly on $$(0,1)$$ but that $$g_n$$ does. I know that it is not something difficult, but I still would like to know if my proofs are correct, please.

First of all, we remark that $$\lim_{n\to\infty}f_n(x)=1$$ and $$\lim_{n\to\infty}g_n(x)=x$$. We prove now that $$f_n$$ doesn't converge uniformly to $$1$$:

$$\bullet$$ First approach: By definition, to show that $$f_n$$ doesn't converge uniformly to $$1$$, we have to satisfy the following: $$\exists \epsilon>0 \ \forall N \ \exists n\ge N \ \exists x\in (0,1)$$: $$|\frac{1}{nx+1}-1|>\epsilon$$. If we take $$x=1/n$$ and $$\epsilon=1/3$$, we satisfy the definition. Therefore, $$f_n(x)$$ doesn't converge uniformly to $$1$$ on $$(0,1)$$

$$\bullet$$ Second approach: Suppose by absurd that $$f_n$$ converges uniformly to $$1$$. Then, we have the following: $$\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$$:

$$|\frac{1}{nx+1}-1|<\epsilon$$

But, $$|\frac{1}{nx+1}-1|=|\frac{-nx}{nx+1}|\le|\frac{nx}{nx}|=1$$. Taking $$\epsilon=2$$ we got the contradiction. So, $$f_n$$ doesn't converge uniformly to $$1$$.

Now, we show that $$g_n$$ converges uniofrmly to $$x$$ on $$(0,1)$$. By definition we have to satisfy the following: $$\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$$:

$$|\frac{x}{nx+1}-x|<\epsilon$$.

But,

$$|\frac{x}{nx+1}-x|\le|\frac{x}{nx+1}|\le|\frac{x}{nx}|\le|\frac{1}{n}|$$.

If we pick $$N$$ such that $$\frac{1}{N}< \epsilon$$ we obtain that $$f_n$$ converges uniforomly to $$x$$ on $$(0,1)$$

• Both of the claims are false. Both of those functions converge pointwise to zero. Feb 22, 2021 at 19:07
• Don't they both converge to $0$? In this case, your proof that $f_n$ doesn't converge to $1$ uniformly is obvious since it doesn't even converge to $1$ pointwise. For $g_n$, this line isn't true $|\frac{x}{nx+1}-x|\le|\frac{x}{nx+1}|\le|\frac{x}{nx}|\le|\frac{1}{n}|$ Feb 22, 2021 at 19:07

Both your pointwise limits for $$f_n$$ and $$g_n$$ are wrong.

Hint.

For every $$x\in (0,1)$$, $$\lim_{n\to\infty}\frac{1}{nx+1}=0,\quad \lim_{n\to\infty}\frac{x}{nx+1}=x\lim_{n\to\infty}\frac{1}{nx+1}=0$$

So both $$f_n$$ and $$g_n$$ converges to $$0$$ pointwise on $$(0,1)$$.

To show that $$f_n$$ does not converge uniformly to zero, consider the sequence $$x_n=\frac1n$$ and estimate $$|f_n(x_n)-f(x_n)|$$ It is easy to see that this quantity does not go to zero.

For $$g_n$$, consider the estimate $$|\frac{x}{nx+1}|=|\frac{1}{n+\frac{1}{x}}|\le\frac1n$$

hint

You made a mistake in your $$\lim_{n\to+\infty}f_n(x)=1$$.

In fact, $$\lim_{n\to +\infty}f_n(x)=0$$

and $$M_n=\sup_{0

thus, the convergence is not uniform at $$(0,1)$$. Or

$$M_n\ge f_n(\frac 1n)=\frac 12$$

By the same, let $$G_n(x)=|g_n(x)-0|$$ then $$G_n'(x)=\frac{1}{(nx+1)^2}$$ and $$\sup_{0

• Forget, I'm stupid... I confused with $x^n$. Thank you for an answer Feb 22, 2021 at 19:15