Pointwise => uniform convergence when?

[St41n]

If we have that:
$$f_T \left( x \right) \to f\left( x \right)$$ for each x (pointwise convergence)
and also that:
$$f_T$$ and $$f$$ are bounded or/and uniformly continuous functions then can we show that there is also uniform convergence?
If no why not? Can you show it with an example?

In general, are there any cases where pointwise convergence implies uniform convergence? I can't find any proof on that. Can you guide me on this?

Thanks in advance

 

[jostpuur]

Having functions bounded doesn't help with uniform convergence at least.

It is not clear to me what you mean by uniform continuity. Do you mean that each $f_T$, with fixed T, is uniformly continuous so that the choice of $\delta$, with given $\epsilon$, doesn't depend on the point $x$? If so, it doesn't help with the uniform convergence.

If you instead meant that the collection of all $f_T$ is uniformly continuous so that the choice of $\delta$ doesn't depend on T, then I'm not sure right now.

 

[St41n]

I actually meant the first thing you said.
So, there is no known way where pointwise implies uniform convergence?

 

[jostpuur]

First consider this example:

$$f_N:[0,1]\to\mathbb{R}, \quad f_N(x) = \left\{\begin{array}{ll} 1,\quad &\frac{1}{N^2} \leq x \leq \frac{1}{N}\\ 0,\quad &0\leq x < \frac{1}{N^2}\;\textrm{or}\;\frac{1}{N} < x \leq 1\\ \end{array}\right.$$

Now $$\lim_{N\to \infty}f_N(x)\to 0$$ for all $$x\in [0,1]$$, but $$f_N$$ do not converge to zero uniformly.

This was not an example with continuous functions, but it is easy to see that you can smoothen these functions a little bit so that each of them alone becomes uniformly continuous, and still the main effect remains the same. This way you get a sequence of continuous, and uniformly bounded functions, which converge towards a continuous function, but still the convergence is not uniform.

 

[jostpuur]

It could be that this kind of claim is right:

Let $$f_N:X\to Y$$ be a sequence of continuous functions between two metric spaces, of which $$X$$ is compact. If $$f_N\to f$$ pointwisely, and if the sequence is uniformly continuous so that for all $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that

$$f_N(B(x,\delta))\subset B(f_N(x),\epsilon)\quad\forall x\in X,\; \forall N\in\mathbb{N},$$

then also $$f$$ is continuous, and $$f_N\to f$$ uniformly.

(If that's not right, then it could be that something similar looking is still right)

 

[St41n]

First consider this example:

$$f_N:[0,1]\to\mathbb{R}, \quad f_N(x) = \left\{\begin{array}{ll} 1,\quad &\frac{1}{N^2} \leq x \leq \frac{1}{N}\\ 0,\quad &0\leq x < \frac{1}{N^2}\;\textrm{or}\;\frac{1}{N} < x \leq 1\\ \end{array}\right.$$

Now $$\lim_{N\to \infty}f_N(x)\to 0$$ for all $$x\in [0,1]$$, but $$f_N$$ do not converge to zero uniformly.

This was not an example with continuous functions, but it is easy to see that you can smoothen these functions a little bit so that each of them alone becomes uniformly continuous, and still the main effect remains the same. This way you get a sequence of continuous, and uniformly bounded functions, which converge towards a continuous function, but still the convergence is not uniform.

That's a nice example thank you, I see what you mean

It could be that this kind of claim is right:

Let $$f_N:X\to Y$$ be a sequence of continuous functions between two metric spaces, of which $$X$$ is compact. If $$f_N\to f$$ pointwisely, and if the sequence is uniformly continuous so that for all $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that

$$f_N(B(x,\delta))\subset B(f_N(x),\epsilon)\quad\forall x\in X,\; \forall N\in\mathbb{N},$$

then also $$f$$ is continuous, and $$f_N\to f$$ uniformly.

(If that's not right, then it could be that something similar looking is still right)

I will take a closer look at this and try to see if I can use it in my case and post any any questions I might have.
Thanks again