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.


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

  • $\begingroup$ 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

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 $$ \begin{aligned} \max|f(x)|&\leq|f(a)|+\max|f'(x)| (b-a)+|f(a)|(b-a)+\max|f'(x)|\\ &=(|f(a)|+\max|f'(x)|) (1+b-a) \\ &=||f||(1+b-a) . \end{aligned} $$


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.