Who's up to the challenge? 14

Calculus Level 5

$f,g$ are real functions that are integrable on $[0,1]$ . If $\displaystyle\int_0^1 |f(x)|^7 \, dx = 1$ and $\displaystyle\int_0^1 |g(x)|^7 \, dx = 16384,$ find the maximum value of $\displaystyle\int_0^1 |f(x)+g(x)|^7 \, dx.$

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The answer is 78125.

1 solution

Wing Tang
Feb 22, 2016

Using H $\ddot o$ lder's inequality and binomial theorem alternatively, we will obtain the answer. Steps are shown as follows.

$\begin{aligned} \int^1_0 |f(x) + g(x)|^7 dx &\leq \int^1_0 \left(|f(x)| + |g(x)|\right)^7 dx\\ &= \int^1_0 \sum^7_{r=0} {7 \choose r}|f(x)|^r |g(x)|^{7-r} dx\\ &= \sum^7_{r=0} {7 \choose r} \int^1_0 |f(x)|^r |g(x)|^{7-r} dx\\ &\leq \sum^7_{r=0} {7 \choose r} \left(\int^1_0 |f(x)|^7\right)^{\frac{r}{7}} \left(\int^1_0 |g(x)|^7 dx\right)^{\frac{7-r}{7}}\\ &= \sum^7_{r=0} {7 \choose r} (1)^{\frac{r}{7}} \cdot \left(16384\right)^{\frac{7-r}{7}}\\ &= \sum^7_{r=0} {7 \choose r} \left( ^7\sqrt{16384}\right)^{7-r}\\ &=\sum^7_{r=0} {7 \choose r} \left(4\right)^{7-r}\\ &= \left(1+ 4\right)^7 = 78125 \end{aligned}$

nice solution!

the intended solution was to use minkowski's inequality,which easily follows from Holder's inequality

you can look at my proof of it here

Holder's inequality

Hamza A - 5 years, 3 months ago

Who's up to the challenge? 14

The answer is 78125.

1 solution

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