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Let $g:\mathbb{R}^n \rightarrow \mathbb{R}$, let $t \in \mathbb{R}$, suppose $s \in [0, t]$ and let $e_i$ denote the standard $i^{th}$ basis vector. I have read the following claim:

$$ \frac{1}{t} \int^t_0 |g(x + se_i) - g(x)|ds \leq \max_{0 \leq s \leq t}|g(x + se_i) - g(x)| $$

How can I see that this claim is true? I really have no thoughts as to how one might proceed to show this except, perhaps, to convert the LHS to a Riemann sum and manipulate that to show that it is less than the RHS. But this does not seem to be a very good approach.

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2 Answers 2

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$$ \begin{aligned} \frac{1}{t} \int^t_0 |g(x + se_i) - g(x)|ds &\leq \max_{0 \leq s \leq t}|g(x + se_i) - g(x)| \times \frac{1}{t}\int_0^t \, ds \\ &= \max_{0 \leq s \leq t}|g(x + se_i) - g(x)| \end{aligned} $$

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  • $\begingroup$ Sorry, but this doesn't help. The key to my understanding this problem is knowing where the "max" term comes from. $\endgroup$ Jun 26, 2012 at 12:36
  • $\begingroup$ Is there a theorem or some other fact that you're using to derive the first inequality? $\endgroup$ Jun 26, 2012 at 12:37
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    $\begingroup$ Assume $f(x) \leq M$ for every $x \in [a,b]$. Then $$\left| \int_a^b f(x)\, dx \right| \leq M (b-a).$$ $\endgroup$
    – Siminore
    Jun 26, 2012 at 12:42
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Substitute $s:=\tau\, t$ $\ (0\leq \tau\leq 1)$ in the integral. Then you get $$ \frac{1}{t} \int^t_0 |g(x + se_i) - g(x)|ds =\int_0^1 |g(x + \tau\, t e_i) - g(x)|\ d\tau\ ,$$ which is obviously $\ \leq \max_{0 \leq s \leq t}|g(x + se_i) - g(x)|$.

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