Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have a set $S$ which contains all functions $v \in C^{1}[0,1]$ so that $v(0) = 0$ and

$$\int_{0}^{1} |v'(x)|^2 dx = 1.$$

How can I show that $S$ is bounded with the infinity norm. That is, how can I show that

$$ sup_{v\in S}\lVert v\rVert_{\infty} $$


I know that for each $v \in S$ there exists a constant $M(v)$ so that $|v(x)| \le M(v)$ for all $x \in [0,1]$ since $v$ is continuous on a closed interval $[0,1]$. What I fail to see is how $\int_{0}^{1} |v'(x)|^2 dx = 1$ helps me.

share|cite|improve this question
Assume you have $v_n=n\cdot x$. Then $v_n\in C^1[0,1]$ and $v_n(0)=0$ is fulfilled for all $n$. But $\sup_{v\in S}||v||_{\infty} \geq \sup_{n}||v_n||_{\infty}\rightarrow \infty$. – k1next Dec 17 '12 at 20:24
@macydanim, so you are saying that $sup_{v\in S}\lVert v\rVert_{\infty} = \infty$ and thus $S$ is not bounded? – user53477 Dec 17 '12 at 20:27
I have no idea what @macydanim is trying to say. Because $v_n(x) = nx$ certainly doesn't satisfy $\int_0^1 |v_n'(x)|^2 \,dx = 1$. – kahen Dec 17 '12 at 20:28
@kahen , exactly. I just wanted to provide an example that shows that the condition $\int_0^1 ||v'|| dx$ is important and what can go wrong otherwise. Sorry if that did not come through. – k1next Dec 17 '12 at 20:30
@macydanim, so is there something very obvious that I might be missing from that integral? – user53477 Dec 17 '12 at 20:35
up vote 2 down vote accepted

Let $v\in S$ and $x\in[0,1]$. Then $$ |v(x)|=\Bigl|\int_0^xv'(t)\,dt\Bigr|\le\int_0^x|v'(t)|\,dt\le\Bigl(\int_0^x|v'(t)|^2dt\Bigr)^{1/2}\Bigl(\int_0^xdt\Bigr)^{1/2}\le\sqrt x\le1. $$ Use has been made of the Cauchy-Schwarz inequality and the fact that $$ \int_0^x|v'(t)|^2dt\le\int_0^1|v'(t)|^2dt\le1. $$

share|cite|improve this answer
Thank you. I'm still wondering why I couldn't see that connection. – user53477 Dec 17 '12 at 21:05
From your expression we have that $-\sqrt{x} \le v(x) \le \sqrt{x}$. So $\sqrt{x} \in S$. But now we have a problem since $\int_{0}^{1} |\frac{d}{dx} \sqrt{x}|^2 dx = \int_{0}^{1} \frac{1}{2 \sqrt{x}}^2 dx = \int_{0}^{1} \frac{1}{4x} dx$ which does not converge. Do we really have $v(x) \le \sqrt{x}$? or perhaps $v(x) \lt \sqrt{x}$? – user53477 Dec 17 '12 at 22:57
We certainly have $|v(x)|\le\sqrt x$. Can you spot something wrong in the proof? But this does nor mean that $\sqrt x\in S$. – Julián Aguirre Dec 18 '12 at 10:03

Another one by contradiction.

Suppose $\sup_{v\in S} || v||_{\infty} =\infty$ That means there exists a sequence $(x_n)\rightarrow \tilde{x}>0$ such that $v(x_n)\rightarrow \infty$. Now the mean value theorem states, that $\exists \xi $ such that $\xi \in [0,{x_n}]$ and $$ v'(\xi) = \frac{v(x_n)-v(0)}{{x_n} -0}=\frac{v({x_n})}{{x_n}} $$ So we know
$$ |v'(\xi)| =|\frac{v({x_n})}{{x_n}}| \rightarrow |\frac{v(\tilde{x})}{\tilde{x}}|= \infty\, \text{ as } n \rightarrow \infty$$ Now as $v\in C^1[0,1]$ we know that $v'\in C^0[0,1]$ so $\int_0^1 || v'(x) || dx \rightarrow \infty$. Which is a contradiction.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.