# Convergence of Sequence in C[0,1] Norms

• BSCowboy
In summary, the statement that if a sequence \{f_n\} is convergent in \left(C[0,1],||\cdot||_{\infty}\right) then it is also convergent in \left(C[0,1],||\cdot||_1\right) is true because the supremum norm is the greatest value in all p-norms. This means that if a sequence is convergent in the supremum norm, it will also be convergent in all other norms on the same space. This is because the supremum norm being finite implies that all other norms are also finite. Additionally, the fact that a sequence converges means that for any given epsilon, there exists a value N

#### BSCowboy

If a sequence $$\{f_n\}$$ is convergent in $$\left(C[0,1],||\cdot||_{\infty}\right)$$ then it is also convergent in $$\left(C[0,1],||\cdot||_1\right)$$.

I think I understand why this is true. (In my own words) The relationship between the supremum norm and the usual norm (really any p-norm) is that the supremum norm is the greatest value in all p-norms. So, for all sequences, if the sequence is convergent in the supremum norm it's convergent in all norms on the same space. Is this true?

Also, for a sequence to converge it means
$$\exists \, f \,\ni \,\forall \,\epsilon>0 \,\exists\, N\ni\, \forall\, x\in C[0,1]$$
$$||f_n-f||_{\infty}<\epsilon \quad \forall n>N$$

This is a given, but how could I use that to prove the implied part? This is for my own edification.

Also, I can think of a counter example to show the other direction is not true.
Such as, $$f_n(t)=t^n \quad \text{then} \quad ||f_n||_1\rightarrow 0$$
but, $$||f_n||_{\infty}\rightarrow 1$$

The first statement is not true in general, only for measure spaces where the total measure is finite.

Counterexample on the real line:

fn(x)=1 -n<x<n, =0 |x|>n+1, and connect up to be continuous in between.
sup |fn(x)|=1, while all Lp norms become infinite.

Right, thank you. That is a good point. I haven't yet thought about this proclamation in all spaces. My statement was about the behavior was concerning this normed metric space.

Since (C[0,1],$$||\cdot||_{\infty}$$) is complete and $$\{f_n\}$$ is convergent,
we know every sequence in (C[0,1],$$||\cdot||_{\infty}$$) is Cauchy convergent and converges uniformly $$\Rightarrow \quad ||f_n-f||_{\infty}\rightarrow \, 0$$.
Because of this we also know:
$$||f_n-f||_1=\int_0^1|f_n(t)-f(t)|dt\leq\int_0^1||f_n-f||_{\infty}dt=||f_n-f||_{\infty}$$
Therefore,
$$||f_n-f||_1\rightarrow \, 0$$

Is my reasoning correct?

Yes, You can use the same idea for all Lp norms.