- #1

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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

[tex]\exists \, f \,\ni \,\forall \,\epsilon>0 \,\exists\, N\ni\, \forall\, x\in C[0,1][/tex]

[tex]||f_n-f||_{\infty}<\epsilon \quad \forall n>N[/tex]

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, [tex]f_n(t)=t^n \quad \text{then} \quad ||f_n||_1\rightarrow 0[/tex]

but, [tex]||f_n||_{\infty}\rightarrow 1[/tex]