# Show that the two norms in $E=\left\{f\in \mathcal{C}^1[0,1]:f(0)=0\right\}$ are equivalent?

I can't show that the two norms defined as $$||f||_{\infty}=\sup_{x\in[0,1]}|f(x)+ f'(x)|$$ and $$N(f)=\sup_{x\in[0,1]}|f(x)| + \sup_{x\in[0,1]}|f'(x)|$$ are equivalent in $$E=\{f\in\mathcal{C}^1([0,1]) \text{ s.t. } f(0)=0\}$$.

A first inequality in one sense is trivial.

For the other inequality, I can write:

$$f(x)=f(0)+\int_0^x f'(t) dt$$ Which implies $$\sup_{x\in[0,1]}|f(x)| \leq \sup_{x\in[0,1]}|f'(x)| .$$ And then $$N(f)\leq 2 \sup_{x\in[0,1]}|f'(x)| .$$ But I can't conclude from here.

I sincerely thank you for your help.

• Thank you for your correction. Commented Jan 5, 2019 at 9:58
• @PaulFrost it does not satisfy $f(0)=0$. Commented Jan 5, 2019 at 11:07

As you have already shown, for $$f \in C^{1}[0,1]$$ such that $$f(0) = 0$$, $$\sup_{x \in [0,1]} |f(x)| \le \sup_{x \in [0,1]} |f'(x)|. \tag{1}$$

We can use this to show that $$\sup |f|$$ is bounded by a constant times $$\sup |f + f'|$$, as follows: \begin{align*} \sup_{x \in [0,1]} |f(x)| &\le \sup_{x \in [0,1]} |e^x f(x)| \\ &\le \sup_{x \in [0,1]} \left| \frac{d}{dx} e^x f(x) \right| \qquad\qquad \text{(by (1), since e^0 f(0) = 0})\\ &= \sup_{x \in [0,1]} |e^x f(x) + e^x f'(x)| \\ &\le e \sup_{x \in [0,1]} |f(x) + f'(x)|. \end{align*}

Now apply triangle inequality to $$|f'| = |f' + f - f|$$: \begin{align*} \sup |f'(x)| &= \sup |f'(x) + f(x) - f(x)| \\ &\le \sup |f(x) + f'(x)| + \sup |f(x)| \\ &\le \sup |f(x) + f'(x)| + e \cdot \sup |f(x) + f'(x)| \\ &= (e + 1) \sup |f(x) + f'(x)|. \end{align*}

Therefore, $$\sup |f(x)|$$ and $$\sup |f'(x)|$$ are both bounded above by a constant times $$\sup |f(x) + f'(x)|$$. This shows that your two norms are equivalent.

Two metrics on the same set are equivalent (i.e. generate the same topology ) iff they have the same set of convergent sequences, because the closure operator in a metric space is entirely and uniquely determined by the convergent sequences.

For brevity let $$\sup_{x\in [0,1]}|g(x)|=\|g\|_S$$ for any continuous $$g:[0,1]\to \Bbb R.$$

Idea: Consider that $$f(x)+f'(x)=e^{-x}(e^xf(x))'.$$

(1). If $$\lim_{n\to \infty}\|f_n-f\|_{\infty}=0$$:

Then $$\lim_{n\to \infty}\|e^{-x}(e^x (f_n(x)-f(x))'\|_S=0,$$ which implies that $$\lim_{n\to \infty}\|(e^x(f_n(x)-f(x))'\|_S=0,$$ which, since $$f_n(0)=f(0)=0,$$ implies that $$\lim_{n\to \infty}\|e^x(f_n(x)-f(x))\|_S= \lim_{n\to \infty}\sup_{x\in [0,1]} |\int_0^x(e^t(f_n(t)-f(t))'dt\,|=0$$ which implies that $$\lim_{n\to \infty}\|f_n-f\|_S=0.$$ Now $$\|f'_n-f'\|_S\leq \|(f_n+f'_n)-(f+f')\|_S+\|f-f_n)\|_S=\|f_n-f\|_{\infty}+\|f_n-f\|_S,$$ so we have $$\lim_{n\to \infty}\|f'_n-f'\|_S=0.$$ Since $$N(f_n-f)=\|f_n-f\|_S+\|f'_n-f'\|_S,$$ therefore $$\lim_{n\to \infty}N(f_n-f)=0.$$

(2). If $$\lim_{n\to \infty}N(f_n-f)=0$$:

Since $$N(f_n-f)\geq \|f_n-f\|_{\infty},$$ therefore $$\lim_{n\to \infty}\|f_n-f\|_{\infty}=0.$$

• Equivalence of norms is stronger than this. One needs to find $m, M > 0$ such that $m N(f) \le \|f\|_\infty \le MN(f)$ for all $f$. Commented Jan 5, 2019 at 11:21
• @TheoBendit. If two norms on a normed linear space are topologically equivalent then they are uniformly equivalent, i.e. $m$ and $M$ exist. Which is an easy exercise: If not, suppose $\|v_n\|_{(1)}>n\|v_n\|_{(2)}$ for each $n\in \Bbb N.$ Let $w_n=\frac {v_n}{\|v_n\|_{(1)}}.$ Then the sequence $(w_n)_{n\in \Bbb N}$ converges to $0$ with respect to $\|\cdot\|_{(2)}$ but $\|w_n\|_{(1)}=1$. Commented Jan 5, 2019 at 12:19