The convergence of different metrics on the same space The following example is from my notes, and I would like clarification on some wider points connected to it, namely about extensions from what we understand metrics and metric spaces to be. It follows on from a previous question that I asked here: Changing of the limits of integration with the integral metric.
We have the integral metric on the space of continuous functions, $C([0,1])$,
$$d_{int}(f,g)=\int_{0}^1 |f(x)-g(x)|\,dx$$
And we are interested in the sequence of functions,
$$f_n(x) =
\begin{cases}
nx  & \text{for $0\le x \le \frac1n$} \\
1 & \text{for $x\ge \frac1n$}
\end{cases}$$
Setting $f(x):=1$, we move to show that $f_n\to f$ in $(C([0,1]),d_{int})$:
$$d_{int}(f_n,f)=\int_{0}^1 |f_n(x)-1|\,dx$$
$$=\int_{0}^{1/n} (1-nx)\,dx$$
$$=\frac1n-n\frac1{2n^2}$$
$$=\frac1{2n}$$
Thus, $d_{int}(f_n,f)\to 0$ as $n \to \infty$, and so $f_n \to f$ in $(C([0,1]),d_{int})$ as we wanted to show.
However, armed with a different metric, namely the uniform metric $d_u$, we notice that $f_n \not\to f$ in $(C([0,1]),d_{u})$.
The uniform metric on the interval $[0,1]$ is defined as,
$$d_u(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$$
Now in my notes it says that since $f_n(0)=0$, it follows that,
$$\sup_{x\in [0,1]}|f_n(x)-f(x)|\ge|f_n(0)-f(0)|=1$$
So then, $d_u(f_n,f)\not\to0$ as $n\to0$.
Here are the points I am finding hard to come to grips with:


*

*What exactly do the integral metric and the uniform metric tell us about a space? I understand that metrics are notions of distance on a space, and understand the examples of the taxi-cab (Manhattan) metric and the Euclidean metric and can visualize them and so see how they are different and what they tell us. But what about the integral and uniform metrics; how can I visualize them and see how they are different? Am I correct in saying that the uniform metric returns the largest distance achieved between two functions?

*As seen in the example above, our sequence of functions $f_n$ converged to $f$ in $(C([0,1]),d_{inf})$ however, $f_n$ did not converge to $f$ in $(C([0,1])d_u)$. How is it that convergence is relative to the metric on the space? And furthermore, what is it that these convergences tells us, depending on the metric used?
 A: The integral (also known as tha $L^1$ metric) and the uniform metric cannot be defined on any space. The uniform metric is defined on sets of bounded, real (or complex) valued functions defined on a set $X$. If $f,g\colon X\to\mathbb{R}$ are bounded, then
$$
d_\text{u}(f,g)=\max_{x\in X}|f(x)-g(x)|.
$$
It measures the greates difference between the values of the two functions.
For the integral metric to be defined, there must be a notion of integral defined on $X$, that is, $X$ shouldbe a measure space. In your example $X=[0,1]$, the measure is Lebesgue measure and
$$
d_\text{int}(f,g)=\int_0^1|f(x)-g(x)|\,dx.
$$
It measures the area between the graphs of the functions $f$ and $g$.
It is clear that $d_\text{int}(f,g)\le d_\text{u}(f,g)$. Thus, if $f_n$ converges to $f$ in the uniform metric, it also converges in the integral metric. But as your example shows, the converse is not true.The area bounded by the graph of $|f-g|$ can be very small. ant the maximum of $|f-g|$ very large.
Finally, in $\mathbb{R}^n$, both notions are equivalent:
$$
\max_{1\le i\le n}|x_i-y_i|\le\sum_{i=1}^n|x_i-y_i|\le n\max_{1\le i\le n}|x_i-y_i|.
$$
