Prove that $X$ is complete but not inner product, and vice versa Let $X$ be the space $C[0, 1]$ under the norm $||·||_{p}$ for $1 \leq p \leq \infty$.
(a) Show that $X$ is complete for $p = \infty$, but it is then not an inner product space.
(b) Show that $X$ is an inner product space for $p = 2$, but it is then not complete.
I thought about this problem for a while, but still got nowhere. Can anybody please help?
 A: *

*The fact that $(C[0,1],\| \cdot \|_{L^\infty})$ is complete follows from the triangle inequality estimate $|f(x)-f(y)| \leq |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)|$. 

*The fact that $(C[0,1],\| \cdot \|_{L^\infty})$ does not have a compatible inner product can be checked by demonstrating that the parallelogram law is not satisfied. One way to do this is to consider two functions $f,g$ whose supports are disjoint. This example actually shows two interesting geometric properties of inner product spaces. First, we can only have $\| x + y \| = \| x - y \|$ if $x$ and $y$ are orthogonal. Second, because of the Pythagorean theorem, we can only have $\| x \| = \| x + y \| = \| x - y \|$ if $y=0$.

*The fact that $(C[0,1],\| \cdot \|_{L^2})$ has a compatible inner product is easy enough to check. The only difficulty is showing that any nonzero continuous function has nonzero square integral, but this can be done by showing that if a continuous function is not zero at a point then it is not zero on an interval around the point.

*The fact that $(C[0,1],\| \cdot \|_{L^2}$ is not complete can be shown by considering a sequence of bounded continuous functions which converge pointwise to a discontinuous function. The boundedness allows the application of the dominated convergence theorem, which ensures the convergence in the sense of $L^2$. To be precise here, we actually need a limit which has no continuous representative in $L^2$. For example, $f_n(x)=x^n$ actually converges in the sense of $L^2$ to the zero function, even though this is not the pointwise limit, so this does not furnish a counterexample. A sequence whose pointwise limit is the indicator function of $[1/2,1]$ would work, however.
Clarifications:


*

*If $f_n$ are continuous and Cauchy in $L^\infty$, then by the completeness of the real numbers, there exists $f$ such that $f_n \to f$ pointwise. Now you need to show that $f_n \to f$ in $L^\infty$ and that $f$ is continuous. Given that $f_n \to f$ in $L^\infty$ (which I won't show here), we can conclude that $f$ is continuous by using the inequality above. In particular we need the uniformity to be able to choose the same $n$ at both $x$ and $y$. 

*If $\| \cdot \|$ is a norm induced by an inner product, then we have the parallelogram law, $2 \| x \|^2 + 2 \| y \|^2 = \| x + y \|^2 + \| x - y \|^2$. Now suppose $f$ and $g$ have disjoint supports. Then $\| f + g \|_{L^\infty} = \| f - g \|_{L^\infty} = \max \left \{ \| f \|_{L^\infty}, \| g \|_{L^\infty} \right \}$. Check that this contradicts the parallelogram law. 

*You should already know that $\| \cdot \|_{L^2}$ is induced by $(f,g) = \int_0^1 f(x) g(x) dx$. So all you need to show is that this map is an inner product. Bilinearity is easy to check using properties of the integral. So the only issue is positive definiteness. So suppose $f$ is not identically zero, then there are $x,\varepsilon$ such that $f(x)^2 > \varepsilon$, which means there is an interval $(x-\delta,x+\delta)$, in which $f(y)^2 > \varepsilon/2$. So $\| f \|_{L^2} > (\varepsilon/2)(2 \delta) = \varepsilon \delta > 0$. Finally $\| 0 \|_{L^2} = 0$ is trivial.

*To find a sequence of continuous functions which converge to the indicator function of $[1/2,1]$, consider functions $f_n$ which are $0$ on $[0,1/2-1/2n]$, $1$ on $[1/2+1/2n,1]$, and increase linearly from $0$ to $1$ on $[1/2-1/2n,1/2+1/2n]$. If you have the dominated convergence theorem, then the convergence desired here is an easy consequence. If you don't, you can do everything explicitly in this simple case and wind up just dealing with a limit of real numbers. 
It remains to show that there is no continuous function which is equal almost everywhere to this indicator function. For this, if $g$ is equal a.e. to this indicator function, then for any $\delta > 0$ we have $x \in (1/2-\delta,1/2)$ with $g(x)=0$ and $y \in (1/2,1/2+\delta)$ with $g(y)=1$. (Why?) Such a $g$ can't be continuous at $1/2$ because its left and right limits disagree there or fail to exist.
A: For part a) we want to show that if some sequence is Cauchy then it converges. We will do this in two steps: first we will show that our sequence is bounded; then we will say that our sequence is really just a bounded sequence of real numbers and invoke that every bounded sequence in $\mathbb{R}$ has a convergent subsequence to show the completeness of $C([0,1])$.
So take some sequence $f_k \in C([0,1])$ and let $\varepsilon > 0$. Suppose that our sequence is Cauchy which is to say that there exists some big $N \in \mathbb{N}$ such that if $n,m \ge N$ then:
$$\sup \limits_{x \in[0,1]}|f_n(x) - f_m(x)| < \varepsilon.$$
Now the supremum is the biggest the distance could be so we may write:
$$|f_n(x) - f_m(x)| < \varepsilon,$$
$$f_m - \varepsilon < f_n < f_m + \varepsilon.$$
We know that $f_k$ is continuous for all $k$. Furthermore a continuous function on a bounded interval is itself bounded. Writing this in symbols we have that $|f_k| < B_k$ for some large bound $B_k$. Thus in particular, taking $k=m$ we can say:
$$-B_m < f_m < B_m$$
and so therefore we have bounded our sequence as:
$$f_m - \varepsilon < f_n < f_m + \varepsilon,$$
$$-B_m - \varepsilon < f_n < B_m + \varepsilon,$$
$$|f_n| < B_m + \varepsilon.$$
Thus,
$$\sup |f_n| < B_m + \varepsilon.$$
So from the above we can say that $|f_n(x)| < B$ for some $B$. In particular fix $x_0$ then the real sequence $f_n(x_0)$ is bounded and it is also Cauchy by hypothesis and so it must converge to some limit call it $f(x_0)$. Thus we have found that for all $x_0$ there must exist an $N$ such that for $n \ge N$ we have $|f_n(x_0) - f(x_0)| < \varepsilon$.
Now $$|f_n(x) - f(x)| \le  |f_n(x) - f_N(x)| + |f_N(x) - f(x)|$$
We know that we can make both of the right hand terms small so that:
$$|f_n(x) - f(x)| \le  |f_n(x) - f_N(x)| + |f_N(x) - f(x)| < \varepsilon.$$
Thus, 
$$\sup|f_n - f(x)| < \varepsilon.$$
Lastly we have to check that the function $f(x)$ is continuous, otherwise $C([0,1])$ wouldn't be complete.
$$|f(x) - f(y)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)|$$
As each $f_n$ is continuous we can make the middle term small by taking $|x-y| < \delta$ and the other two terms can be made small as $f$ is the pointwise limit of $f_n$. Thus $f(x) \in C([0,1])$.
To show the space is not complete you can try the sequence of continuous functions:
$$f_n(x) = \left\{ \begin{array}{c c}1 & 0 \le x \le 1/2 - 1/n\\
-nx + n/2 &1/2- 1/n \le x \le 1/2 + 1/n\\
-1 & 1/2 + 1/n \le x \le 1 \end{array} \right.$$
Check that this sequence is Cauchy and then show it doesn't converge in $C([0,1])$.
If you draw this function it makes more sense and you will see that what is happening is that you have a continuous function for each $n$ but in the limit as $n$ goes to infinity it becomes discontinuous.
