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I have tried solving this problem.

Show that $$d(f,g)=\int_X |f-g|^p d\mu$$ defines metric space when $0<p\leq 1$.

The only thing I need to show is triangular inequality.

I have tried to use Holder inequality but it does not work since $p\leq 1$. So I think, I need to prove that

$$|f-h|^p\leq |f-g|^p+|g-h|^p$$

directly. Buy I don't know how... seems very easy... but ....

Any hint or answer would be helpful! Thanks in advance

1 Answers1

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Using the triangle inequality, you can expand as:

$$ |f-h|^p = |f-g+g-h|^p \leq (|f-g|+|g-h|)^p $$

And then you can use the result given by this question to get your result.

Sambo
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  • It turned out that the link in this answer was not appropriate to answer my question. Here is a correct link that I found . https://math.stackexchange.com/q/134714/523306 – Byeong-Ho Bahn Aug 29 '18 at 01:26
  • It wasn't? Could you explain why so I could edit my answer? – Sambo Aug 29 '18 at 02:16
  • The function value of $f$ and $g$ can be negative. Your link only deals with positive number. Note that he used the property that $(p+q)^{t-1}\leq p^{t-1}$ which is not true if we switch $(p+q)$ to $|p+q|$ with positive $p$ and negative $q$. – Byeong-Ho Bahn Aug 29 '18 at 11:14
  • Right, and this is why, in my answer, I mentioned that we should first use the triangle inequality. – Sambo Aug 29 '18 at 11:50
  • I was trying to understand based on your argument but I could not, Could you expand your answer a little bit? – Byeong-Ho Bahn Aug 29 '18 at 17:29
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    I suppose, but I'm not sure it'll be helpful. The regular triangle inequality gives us $|f-g+g-h| \leq |f-g| + |g-h|$, and so we know that $|f-g+g-h|^p \leq (|f-g| + |g-h|)^p$. Then if we let $a=|f-g|$ and $b=|g-h|$, both $a$ and $b$ are nonnegative, so we can use the link I provided to get $(a+b)^p \leq a^p + b^p$. Combining this with the above inequality, we get the desired $|f-g+g-h|^p \leq |f-g|^p + |g-h|^p$. – Sambo Aug 30 '18 at 03:31
  • You are right! I think I made a mistake.. Thanks! – Byeong-Ho Bahn Aug 30 '18 at 15:08