For $x\in C^{1}[0,1]$ let: $${\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert x \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_1=\lvert x(0) \rvert + \lVert x'\rVert_{\infty}$$ $${\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert x \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_2=\max \left\{\left|\int_{0}^{1} x(t) \,dt\right|, \lVert x'\rVert_\infty \right\}$$ $${\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert x \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_3= \Big ( \int_{0}^{1} |x(t)|^2 \,dt+ \int_{0}^{1} |x'(t)|^2 \,dt \Big)^{\frac{1}{2}} $$

It is required to find out which of these norms is equivalent to the norm $ {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert x \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}=\lVert x \rVert_\infty + \lVert x' \rVert_\infty.$

To prove that two norms $|x|_1$ and $ |x|_2$ on a normed vector space $V$ are equivalent, one needs to prove the existence of two constants $A$ and $B$ such that $A|x|_1 \leq |x|_2 \leq B|x|_1, \forall x \in V. $ I have tried for some time without a conclusive answer.

Can somebody give me some hint. Thanks.

  • $\begingroup$ $\|x\|_2$ is equivalent to $ |\int_0^1x(t)dt|+\|x'\|_{\infty}$ because for any non-negative real $r, s$ we have $r+s=$ $ \max (r,s)+\min (r,s)\geq$ $ \max (r,s)\geq$ $ (1/2)\max (r,s)+(1/2)\min (r,s)=$ $(r+s)/2.$ $\endgroup$ – DanielWainfleet Nov 17 '17 at 10:30
  • $\begingroup$ A useful fact for 2: by the intermediate value theorem, for any $x$ there exists $t_0$ such that $x(t_0) = \int_0^1 x(t)\,dt$. And you can write $x(t) = x(t_0) + \int_{t_0}^t x'(s)\, ds$ $\endgroup$ – Nate Eldredge Nov 17 '17 at 14:43

The easiest inequalities to prove are: $$ \newcommand{\triple}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \triple{x}_1 \leq \|x\|_\infty + \|x'\|_\infty\\ \triple{x}_2 \leq \|x\|_1 + \|x'\|_\infty \leq \|x\|_\infty + \|x'\|_\infty \\ \triple{x}_3 \leq (\|x\|_\infty^2 + \|x'\|_\infty^2)^2 \leq \|x\|_\infty + \|x'\|_\infty. $$

By the fundamental theorem of calculus we also have that $$ x(t) = x(0) +\int_0^t x'(s) ds, $$ this should give you another inequality between $\triple{x}_1$ and $\triple{x}$, proving that they are equivalent.

Correction due to Nate Eldredge: The second norm $\triple{\cdot}_2$ is also equivalent to $\triple{\cdot}$. To prove this note that the intermediate value theorem implies that there's a $t_0 \in [0,1]$ such that $x(t_0) = \int_0^1 x(t)$. We also have that $$ x(t) = x(t_0) + \int_{t_0}^t x'(s) ds = \int_0^1 x(s) ds + \int_{t_0}^t x'(s)ds. $$ This implies the inequality $\|x\|_\infty \leq \Big \lvert \int_0^1 x(t) dt \Big \rvert + \|x'\|_\infty$, which will give you a second inequality between $\triple{x}_2$ and $\triple{x}$.

The third norm $\triple{\cdot}_3$ is also not equivalent to $\triple{\cdot}$. It might be easier to first find a proof that the $\|x\|_2 = \left(\int_0^1 \lvert x(t)\rvert^2 dt \right)^{1/2}$ isn't equivalent to $\|x\|_\infty$ in $C^0[0,1]$ and then adapt that proof to this case.

  • $\begingroup$ @NateEldredge Of course, I thought I could do something with sine functions but that obviously didn't turn out. I will edit in your other suggestion, thank you. $\endgroup$ – Demophilus Nov 17 '17 at 15:54

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.