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Fix a number $k > 0$ and let $$T(x) = \begin{cases} k &: x \geq k\\ x &: |x| < k\\ -k &: x \leq -k \end{cases}. $$ Define $S(s) = \int_0^s T(|x|^{m-1}x)\;dx.$

I want to show that $$S_k(s) \geq \frac{1}{1+m}|T(|s|^{m-1}s)|^{\frac{m+1}{m}}$$ and $$S_k(s) \leq k|s|.$$

The last one can be done like: $|S_k(s)| \leq |s-0|k$ because the integrand is bounded by $k$.

For the first one I tried to compute $$S(s) = \begin{cases} \frac{1}{m+1}|s|^{m+1} &: |s| \leq k^{\frac 1m}\\ \frac{1}{m+1}|s|^{m+1} + (|s|-k^{\frac 1m})k &: |s| > k^{\frac 1m} \end{cases} $$ (I hope it is correct) but I don't know where to go now. Can someone help me?

(This is extracted from here on page 372).

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up vote 1 down vote accepted

Look at $s\geq 0$:

when $0\leq s\leq k, S_k(s)=\int_0^s x^m dx=\frac{s^{m+1}}{m+1}=\frac{1}{m+1}T(|s^{m-1}|s)^\frac{m+1}{m}$;

when $s\geq k, S_k(s)\geq S_k(k)=\frac{1}{m+1}T(|k^{m-1}|k)^\frac{m+1}{m}=\frac{1}{m+1}T(|s^{m-1}|s)^\frac{m+1}{m}$.

The same way to prove $s\leq 0$.

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Thank You. !!!!! – delimit Jul 22 '14 at 17:01

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