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.


(i) (1) follows because the sum of two norms is a norm. (2) is not a norm; $\|f'\|_\infty = 0 $ does not imply $ f \equiv 0.$ (3) is a norm, even though it the sum of a norm and a non-norm. (4) I do not understand your question "What is $|f(0)|?$" Anyway, this is a norm: Show $|f(0)| +\|f'\|_1 = 0 \implies f' \equiv 0 \implies$ $f$ is constant; the condition $|f(0)| =0$ then nails down the constant as $0.$

(ii) I don't know what this is about either. However, $\|f\|_1,\|f\|_\infty$ are not equivalent.



  1. $\|\cdot\|_A$ is a norm since it is a sum of norms;
  2. $\|\cdot\|_B$ 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;
  3. 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 $\|\cdot\|_\infty$;
  4. 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 $f\equiv0$.


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.


To prove that these are all norms all you must do is prove absolute homogeneity and show that they satisfy the triangle inequality. Then separation of points follows directly from this.

With regards to part (ii) I will also be no help unfortunately...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.