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Given the space of continuous functions on the closed interval $(a,b)$ to $K$ and defining the supremum norm on this space one defines the infinite normed linear space $K$. More then that it is a complete space concerning this norm and thus a Banach space.

I have two questions: - I would like to prove that it is complete using the supremum norm and it is not a complete space in respect to the $p$-integral norm. - beyond this question how one usually determines the norm of a linear mapping, i.e. an $n\times n$ matrix on a concrete example (it is defined as the supremum of the values of the matrix on unit vectors).

Thanks for your comments.

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ivo
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2 Answers2

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For the first question, suppose $f_n$ is Cauchy in the uniform norm. Then $f_n$ is also Cauchy at each point, so because the reals are complete, $f_n$ is pointwise convergent. Call its pointwise limit $f$. You want to show $f$ is continuous. Use the triangle inequality:

$$|f(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$$

for any $n$. Try to bound the right side by choosing $x$ close to $y$ and $n$ large. (Spend some time thinking about this argument, if you haven't seen one like it before. Small variations on this setup are useful in a lot of different contexts.)

For the second question, you have to find a sequence of continuous functions which is Cauchy in $L^p$ but does not converge to any continuous function in $L^p$. This is harder than it sounds, because a sequence of functions can converge in $L^p$ to something which is not its pointwise limit. This means that you cannot simply argue that "the sequence converges in $L^p$ to a discontinuous function, therefore it cannot converge in $L^p$ to a continuous function".

So it is easiest to work with a concrete example. The idea is to take a sequence whose pointwise limit has a jump. For instance, you can take functions $f_n$ which are $1$ on $[0,1/2-1/n]$,$0$ on $[1/2+1/n,1]$, and linear on $(1/2-1/n,1/2+1/n)$, where $n \geq 2$. Assume that $f$ is a continuous function on $[0,1]$ and find $\delta>0$ so that $|f(x)-f(y)|<1/4$ if $|x-y|<\delta$. Then argue that $\int_{1/2-\delta}^{1/2+\delta} |f_n-f|^p dx$ is bounded below.

Ian
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Firstly using the supremum norm the space of continuous functions you are talking about will be complete if and only if the codomain K is complete.

Since you are assuming K to be a banach space it is already complete so we may proceed to proving that the space of continuous bounded functions is also complete. let X=[a,b], to prove: c(X,K) is complete.

Let {fn} be a cauchy sequence in C(X,K).Then for each 'x' in X,the sequence {fn(x)} is cauchy in Y because e(fn(x),fm(x))<=s(fn,fm) for all m,n in N,and 'e' is the metric on K. Since K is complete {fn(x)} converges in K. We define a function g:X->K by g(x)=lim fn(x) for eack 'x' in X. Since {fn} is bounded and continuous => g is also bounded and continuous and hence g is in C(X,K). Now if we show that fn->g in C(X,Y) we ll be done.

So suppose r>0,because fn is cauchy there exists 'h' in C(X,K) such that the open ball b[h;r/3) includes a tail of {fn}. so for sufficiently large 'n' we have s(h,fn)g(x) for all 'x' in x,it follows that e(h(x),g(x))<=r/3 for all 'x' in X,yielding s(g,h)<=r/3. Since s(g,h)<=r/3 and b[h;r/3) includes a tail of {fn},certainly b[g;r) includes the same tail. But r>0 is arbitrary so this proves that {fn} converges to g in C(X,K). Since {fn} is an arbitrary Cauchy sequence in C(X,K) Hence we conclude that C(X,K) is complete.

MathLover
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