# Norms and equivalence classes question

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.

# (i)

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 $$f\equiv0$$.

# (ii)

• $$\|f\|_1\leq\|f\|_\infty$$ 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.