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If $ f : [a,b] \subset \mathbb{R} \rightarrow \mathbb{R} $ is continuous and differentiable in $(a,b)$, then one can define a norm for such functions as

$$ \|f\| = |f(a)| + \max_{x \in (a,b)} |f ^\prime(x)| $$

I have already proven that it is in fact a norm.

Now I am trying to prove that this inequality holds.

$$ \max_{x \in [a,b]} |f(x)| \leq (1+b-a)\|f\| $$

There is a hint that I should remember that $f(x) = f(a) +\int_{a}^{x}f^\prime(t) dt$

I have tried several approaches with no success.

If I let $\theta = \arg\max|f(x)|$ and $\lambda = \arg\max|f^\prime(x)|$

Then I could use the hint and say that

$$ \|f\| = \|f(a) + \int_{a}^{x}f^\prime(t) dt \| \leq \|f(a)\| + \|\int_{a}^{x}f^\prime(t) dt \| $$

and I could get that that

$$ 0 \leq \|f\| \leq |f(\theta)| + |f^\prime(\lambda)| $$

but now I have $|f(\theta)| $ on the wrong side of the inequality and I can't find a way to move things around so that I could get any close to my goal.

I tried to define a function $g(x) = (1+b-a)f(x)$, and then I would only have to show that $|f(\theta)| \leq \|g\|$, but I can't find a way to do so.

Any hint would be greatly appreciated; specially about how to use the hint I already have.

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We have$$\left|f\left(x\right)\right|=\left|f\left(a\right)+\int_{a}^{x}f'\left(t\right)dt\right|\leq\left|f\left(a\right)\right|+\int_{a}^{x}\left|f'\left(t\right)\right|dt\leq\left|f\left(a\right)\right|+\max_{t\in\left(a,b\right)}\left|f'\left(t\right)\right|\left(b-a\right)\leq\left(1+b-a\right)\left|f\left(a\right)\right|+\max_{t\in\left(a,b\right)}\left|f'\left(t\right)\right|\left(1+b-a\right)=\left(1+b-a\right)\left\Vert f\right\Vert$$ because $b-a>0$ and $1<1+b-a$. So$$\max_{x\in\left[a,b\right]}\left|f\left(x\right)\right|\leq\left(1+b-a\right)\left\Vert f\right\Vert.$$

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  • $\begingroup$ I clearly was on the wrong path there. Thank you so much! I spent the whole day trying to figure that one out! $\endgroup$ – Sofia Ontiveros Mar 17 '15 at 21:20
  • $\begingroup$ @SofiaOntiveros You're welcome ;) $\endgroup$ – Marco Cantarini Mar 17 '15 at 21:21
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Here is another approach:

Using the MVT, we have that for any $x$ we can find a $c$ such that $$f(x)-f(a)=f'(c)(x-a).$$ Rearranging gives: $$f(x)=f(a)+f'(c)(x-a).$$ Taking absolute values and using the triangle inequality $$|f(x)|=|f(a)+f'(c)(x-a)|\leq |f(a)|+|f'(c)|\cdot|(x-a)|$$ then taking the max of everything: $$\begin{aligned} \max|f(x)|&\leq |f(a)|+\max|f'(c)|\cdot\max|(x-a)|\\ &\leq|f(a)|+\max|f'(x)|\cdot (b-a) \\ \end{aligned}$$

This is a much stronger condition. So we add to the right side $|f(a)|(b-a)+\max|f'(x)|$ to get $$ \begin{aligned} \max|f(x)|&\leq|f(a)|+\max|f'(x)| (b-a)+|f(a)|(b-a)+\max|f'(x)|\\ &=(|f(a)|+\max|f'(x)|) (1+b-a) \\ &=||f||(1+b-a) . \end{aligned} $$

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