# Thread: Sequence of functions convergence.

1. ## Sequence of functions convergence.

Alright I am having some trouble with this topic.

Let:

$f_{n}(x) = log(1 + \frac{1}{n(x-1)})$ x belonging to ]1,+infinity[

I have to study pointwise and uniform convergence.
So I'd like to start with the pointwise and here it is:

$\lim_{f_{n}(x) \rightarrow +\infty} = log1 = 0$

So $f_{n}(x) ----> converges pointwise to f(x) = 0$

Now I have to check the uniform convergence, so I'd begin this way:

$sup|f_{n}(x) - f(x)| \rightarrow 0 = sup|f_{n}(x)| \rightarrow 0$

Okay, here is where I am stuck, what do I have to do now? Do I have to take the first derivative respect to x and make it equal to zero? I am very confused. Thank you!

2. ## Re: Sequence of functions convergence.

Your computation of pointwise convergence is correct: for any given $x \in ]1;\infty[$, you have
$\lim_{n \to \infty} f_n(x) = f(x) = 0$
As to the uniform convergence, you need to check the limit
$\lim_{n \to \infty} \sup_{x \in ]1;\infty[} |f_n(x)-f(x)| = \lim_{n \to \infty} \sup_{x \in ]1;\infty[} |f_n(x)|$
What is the supremum here? Clearly, for values of $x$ approaching $1$ from above, the function $f_n(x)$ is unbounded. In fact,
$\lim_{x \downarrow 1} f_n(x) = \infty$
Therefore,
$\sup_{x \in ]1;\infty[} f_n(x) = \infty$
from which we conclude that the convergence is not uniform (but only pointwise).

3. ## Re: Sequence of functions convergence.

What if we get a restriction to an interval such as [a,+infinity[ with a > 1 , is the convergence uniform in that case?

4. ## Re: Sequence of functions convergence.

Yes. In such case, the supremum (maximum) would be attained at $x = a$ because $f_n(x)$ is positive and monotonically decreasing in $x$:
$\sup_{x \in [a;\infty[} f_n(x) = \log\left(1+\frac{1}{n(a-1)}}\right)$
And this expression tends to zero for $n \to \infty$.

5. ## Re: Sequence of functions convergence.

Alright thank you, I understand. So could you (if I am not asking too much) tell me what are the "usual" steps to take ? When do I have to take the first derivative? Thanks

6. ## Re: Sequence of functions convergence.

I wouldn't know of any "usual steps". The procedure is roughly the same every time you need to check uniform convergence: look out for singularities or discontinuities (emerging in the limit $n \to \infty$) and check if they lead to a failure of uniform convergence. I am not sure what you mean by "first derivative". There is no differentiation involved in what we have discussed above.

I recommend you keep in mind the two typical textbook examples where uniform convergence fails:

$f_n \colon ]0;1] \to \mathbb{R}, x \mapsto \begin{cases} 1-nx \quad &\text{for} \ 0

$f_n \colon ]0;1] \to \mathbb{R}, x \mapsto \begin{cases} 1 \quad &\text{for} \ 0

In a similar way, both series fail to be uniformly convergent. That is because, although we are eventually locked down to zero for every $x$ (pointwise convergence), near $0$ there always remains some little jump up to the value 1, which is such that the supremum of $f_n$ is always $1$ and will never go down to zero.

Note that the former function is continuous (like a bezel, or a ramp on the floor), while the latter function is discontinuous (like a step). So in the former case, the failure of uniform convergence goes along with an ever-increasing slope (the "first derivative" you may have meant), while in the latter case, it goes along with a discontinuity which is not resorbed for $n \to \infty$.