Convergence of Sequence in C[0,1] Norms

[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, $$||f_n||_{\infty}\rightarrow 1$$

 

[mathman]

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

Counterexample on the real line:

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

 

[BSCowboy]

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.

 

[BSCowboy]

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?

 

[mathman]

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

 

[BSCowboy]

Thanks, I appreciate your input.