# Prove that $||f|| :=\int^1_0 |f(t)|dt$ is the norm of the linear space $C[0,1]$ of continuous functions of the interval $[0,1]$.

1. Prove that $$||f|| :=\int^1_0 |f(t)|dt$$ is the norm of the linear space C[0,1] of continuous functions of the interval [0,1]. 2.Is this norm consistent with any scalar multiplication? 3.Is the linear space C[0,1] a Banach space with respect to this norm?

My answers: We need to show three properties: positivity, homogeneity, and the triangle inequality.

Positivity: $$||f|| :=\int^1_0 |f(t)|dt$$≥0

This is true since the absolute value of any real number is non-negative.

Homogeneity: $$||af|| :=\int^1_0 |af(t)|dt=|a|\int^1_0 |f(t)|dt=|a|x||f||$$ ​ ∣f(t)∣dt=∣α∣⋅∣∣f∣∣ Here a is a scalar. The constant multiple can be pulled out of the integral.

Triangle Inequality:$$∣∣f+g∣∣=\int^1_0 |f(t)+g(t)∣dt<=\int^1_0 |f(t)∣dt+\int^1_0 |g(t)∣dt=∣∣f∣∣+∣∣g∣∣$$ This follows from the triangle inequality for integrals.

These three properties confirm that $$||f|| :=\int^1_0 |f(t)|dt$$ is a norm on the linear space C [0,1].

2.Yes, this rate is consistent with scalar multiplication because it satisfies the homogeneity condition. The homogeneity condition means that the norm is satisfied with respect to constant (scalar) multiplication. The expression $$(||\alpha f|| = |\alpha| \cdot ||f||)$$ shows that after multiplying the norm of the function (f) by the absolute value of the constant (\alpha) we will get the same rate as well as multiplying the function (f) by this constant. Therefore, this rate is consistent with scalar multiplication.

1. To determine if the linear space (C[0,1]) is a Banach space with respect to this norm, we need to check if it is complete and closed.

The linear space (C[0,1]) with this norm is complete if every Cauchy sequence in this space converges to a limit in the same space. This can be justified using the Arzelà-Ascoli theorem, which describes when a sequence of bounded functions has a convergent subsequence. The combination of Arzelà-Ascoli conditions and the properties of this norm should ensure that every Cauchy sequence has a limit in this space.

The linear space (C[0,1]) is closed if it contains all its limit points. This can be ensured using convergence properties and the concept of function extensions.

Therefore, the answer depends on whether these conditions are satisfied. If they are, then (C[0,1]) with this norm can be considered a Banach space.

Am I correct? If not any help is appreciated.

• Point 3. is confusing. Why do you need to check that it is closed? Also what does completeness in this case have to do with arzela ascoli?
– jd27
Dec 13, 2023 at 10:54
• Does this answer your question? Is $C[0,1]$ a Banach space for the $L^1$ norm? Dec 13, 2023 at 11:22
• The question in point 1 should be "prove that $\dots$ is a norm" (not the norm). The question and answer in point 2 make no sense (they are redundant with one of the properties proved in point 1). Dec 13, 2023 at 11:24
• math.stackexchange.com/tags/solution-verification/info For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof that is in doubt, and why so. Dec 13, 2023 at 11:30

For (1): Another condition for $$||\cdot||$$ to be a norm is being positive definite, i.e. $$||f|| = 0 \iff f = 0$$. Once you check this as well, $$||\cdot||$$ will be a norm by definition.
For (3): There is no need to check if $$C[0,1]$$ is closed, since in a topological space $$X$$, the set $$X$$ will always be closed. Only completeness needs to be shown in order to be a Banach space. But in this case, the normed space is not complete! Look at the following counterexample:
f_n(x) = \begin{align} \begin{cases} nx \hspace{1cm} \text{if} \hspace{0.2cm} 0 \leq x \leq \frac{1}{n} \\ 1 \hspace{1cm} \text{else} \end{cases} \end{align}.
It's easy to see this converges pointwise to $$f(0) = 0$$ and $$f(x) = 1$$ for $$x > 0$$. Now it's not hard to show that if the sequence would converge, then it would converge to this $$f$$, but $$f$$ is not continuous, giving a contradiction.
The problem with using the Arzelà-Ascoli theorem is that not all conditions of this theorem are satisfied. For example, a Caushy sequence $$f_n$$ in $$C[0,1]$$ need not be uniformly bounded (if you want a counterexample, take inspiration from the previous $$f_n$$), nor equicontinuous. Also if a subsequence of a Caushy sequence converges, that does not imply that the whole sequence converges (this does hold however if we take the norm $$||f||_{\infty} = \sup_{x \in [0,1]} |f(x)|$$).