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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.

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  • $\begingroup$ $\|x\|_2$ 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
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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.

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  • $\begingroup$ @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

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