Answer to Question #164984 in Functional Analysis for John

Question #164984
  • Prove that C.V100√n-1
1
Expert's answer
2021-02-25T02:23:19-0500

We have to prove that "C^1[0,1]" is not complete with the norm:

"||f||_{\\infty}=sup_{x \\in [0,1]}|f(x)|"


The right sequence for norm is "f_n=\\sqrt{x+\\dfrac{1}{x}}"


Notice that "n\\in N:f_n\\in C^1[0,1]"


let "f=\\sqrt{x}"

We see that "f_n" converges to fin sup norm in "C[0,1]" , Thus it is cauchy.


"C^1[0,1]" is a supspace of "C[0,1]" and all terms of "(f_n)" are in "C^1[0,1]" ,So


As the "f_n" converges to "C[0,1]"

"\\implies (f_n) \\text{ is Cauchy in } C^1[0,1]."


So Given space is not complete with norm "||f(x)||_{\\infty}=sup_{x\\in[0,1]}|f(x)|"

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