Pointwise vs Uniform Convergence

**Ant** · October 20th 2012, 10:55 AM

I've been trying, with no luck so far, to get my head around the differnce between pointwise and uniform convergence.

The definition which I am using for Uniform convergence is:

We say $f_{m}$ converges to $f$ uniformly on a set $A$ if

$| f(z) - f_{m}(z)| \leq C_{m},\forall z \in A\\$

Where $C_{m}$ are some constants such that $C_{m} \to 0$ as $m \to \infty$

I understand this definition, but what I am having trouble with is seeing how any convergence sequence of function could fail to satisfy this!

Assume $f_{m}(z) \to f(z)$ but not uniformly.

Now we fix $m$ and calculate $| f(z) - f_{m}(z)|$ for each possible choice of $z \in A$ . There must be a value for $z$ which gives the greatest value (i.e. for which $f_{m}(z)$ is furthest away from $f(z)$ . So take this value and denote is $C_m$ for each choice $m$ . Then for each $m$ we have certainly satisfied $| f(z) - f_{m}(z)| \leq C_{m}, \forall z \in A\\$ .

Moreover this sequence $C_{m}$ must converge to 0 as have by assumption that $f_{m}(z) \to f(z)$ . So $f_{m}$ converges to $f$ uniformly on a set $A$ . Contradicting our assumption that it converged but not uniformly!

Clearly there must be something very wrong with the above argument but I can't see what it is.

thanks for any help or explanation!

**TheEmptySet** · October 20th 2012, 11:38 AM

Here is a counter example for you to think about.

$f_n(x)=\begin{cases}1, \text{ if } |x| \le n \\ 0, \text{ if } |x| > n\end{cases}$

Pointwise $f_n(x) \to 1$

Note that $|f(x)-f_n(x)|=1$ for all n.

**Ant** · October 20th 2012, 11:49 AM

Is $f(x) = 1$ above?

Also, is the domain just: $x \in Real numbers$ ?

Thanks

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