TITLE: Norms and equivalence classes question
QUESTION [1 upvotes]: Let $f\in C[0,1]$. Recall two of the norms we considered in class:
  $$\|f\|_\infty = \sup_{t\in[0,1]}|f(t)|, \quad \|f\|_1 = \int_0^1|f(t)|\ \mathsf dt. $$
  Consider the space $C^1([0,1])$ of continuously differentiable functions on $[0,1]$. Define
\begin{align}
\|f\|_A &= \|f\|_\infty + \|f\|_1, \tag 1\\
\|f\|_B &= \|f'\|_\infty, \tag 2\\
\|f\|_C &= \|f\|_\infty + \|f'\|_\infty, \tag3\\
\|f\|_D &= |f(0)| + \|f'\|_1. \tag 4
 \end{align}
(i) Which of these formulae define norms?
(ii) Consider the set of norms you found in part (i) together with the norms $\|\cdot\|_1$, $\|\cdot\|_\infty$. Identify the equivalence classes of thosese norms with respect to our definition of equivalence.

Part (i) I have sort of proven that (1) , (2) and (3) are norms. For the positivity bit I know I need to prove $||f(x)||$ implies $f(x)=0$.
For example $||f||_A = ||f||_{\infty}+||f||_1$. And the rhs separately imply f=0 how do I show it applies to the whole of the rhs?
(4) I think this is a norm but not sure how to show it. What is $|f(0)|$?
(ii)Not sure what it is asking. I know $||f||_1$ and $||f||_{\infty}$ are equivalent.

REPLY [1 votes]: (i)

$\|\cdot\|_A$ is a norm since it is a sum of norms;
$\|\cdot\|_B$ is not a norm because $f'=0$ does not imply $f=0$: take a constant, it has $\|\cdot\|_B$ equal to 0 but it is not 0. Might be a seminorm though;
To prove $\|\cdot\|_C$ is a norm, you must prove it is nonnegative, which is evident; prove it is zero only on the zero function, which is evident since $\|\cdot\|_\infty$ is and the other part is never negative; prove it is homogeneous, and this follows from homogeneity of $\|\cdot\|_\infty$ and linearity of the derivative; and prove triangle inequality which follows from that of $\|\cdot\|_\infty$;
All properties can be proven similarly to 3; I will only prove it is zero only for $f=0$; that is because if the derivative is 0 the function is constant and the norm here being 0 implies $|f(0)|=0$ so $f(0)=0$ and this means $f\equiv0$.

(ii)
I am not sure about this, but I would interpret this as asking which of these are equivalent.

$\|f\|_1\leq\|f\|_\infty$ since you can pull the $\|f\|_\infty$ out of the integral leaving $\int_0^1dx=1$; hence $\|f\|_A\leq2\|f\|_\infty$; but evidently $\|f\|_\infty\leq\|f\|_A$; hence $\|\cdot\|_\infty$ and $\|\cdot\|_A$ are equivalent.
"Norm B", besides not being a norm, could not be equivalent to norm $\infty$ since a bounded function can oscillate very wildly so there is no way $\|f'\|_\infty=\|f\|_B$ could be bounded by $C\|f\|_\infty$.
Norm C belongs to a distinct equivalence class, since again it cannot be equivalent to norm $\infty$; that is by the above: to bound norm C with norm infinity you would have to bound "norm" B by norm infinity, which, as shown above, is impossible.
Let us compare norms C and D. Now norm D is bounded by norm C since $|f(0)|\leq\|f\|_\infty$ and $\|f'\|_1\leq\|f'\|_\infty$. If we consider a function $f_n$ that starts at 0, raises to $n$ in $[0,\frac1n]$, goes back to zero in $[\frac1n,\frac2n]$ and stays there, and then set $g_n(x)=\int_0^1f_n(x)dx$, we get $\mathcal{C}^1$ function whose derivatives have integral constantly one, satisfying $g_n(0)=0$ for all $n$, but with norms infinity diverging; hence, norm C and norm D cannot be equivalent since, while $\|f\|_D\leq\|f\|_C$, there is no $K$ for which $\|f\|_C\leq K\|f\|_D$ for all $f\in\mathcal{C}^1([0,1])$.
Norm infinity and norm 1 are not equivalent since, while $\|f\|_\infty\geq\|f\|_1$, if you consider a smoothing of the $f_n$'s above, their norms infinity are unbounded but their norms 1 are bounded, provided you don't move too much from the $f_n$'s by smoothing.

So we have 4 equivalence classes: $\{\|\cdot\|_\infty,\|\cdot\|_A\}$, and the classes containing only norm C, only norm D and only norm 1.


TITLE: Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(\frac{1}{i})$ (for large enough $i$)
QUESTION [2 upvotes]: Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(\frac{1}{i})$ (for large enough $i$)
i) Suppose $\sum a_i$ converges. Must $f'(0)$ exist?
ii) Suppose $f(0) = f'(0) = 0$. Must $\sum a_i$ converge?
This is question #6 from Spivak Calculus 4th edition. I already showed that, if $\sum a_i$ converges, $f(0) = 0$, and also that, if $f'(0)$ exists and $\sum a_i$ converges, then $f'(0) = 0$, and also that if $f''(0)$ exists and $f'(0) = f(0) = 0$ then $\sum a_i$ converges.

REPLY [0 votes]: Given that you already did 2 all that is left is to do 1. I can construct a function $f(x)$ such that it is not differentiable(let it be piecewise with the same value at 0, but vastly different slope like say -10000) therefore it might not be differentiable.