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.

  • $\begingroup$ Thank you for your correction. $\endgroup$ – Furdzik Jan 5 '19 at 9:58
  • $\begingroup$ @PaulFrost it does not satisfy $f(0)=0$. $\endgroup$ – Lorenzo Quarisa Jan 5 '19 at 11:07

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.$

  • $\begingroup$ 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 $f$. $\endgroup$ – Theo Bendit Jan 5 '19 at 11:21
  • 1
    $\begingroup$ @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 $\|w_n\|_{(1)}=1$. $\endgroup$ – DanielWainfleet Jan 5 '19 at 12:19

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.


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