Cfrgtkky

On $X = C^0big([a,b]big)$, for any $p in mathbb{R}$, $p>1$, we define the $L^p$ norm by,
$$|f|_{L^{p}}:=big(int^{b}_{a}|f(x)|^{p}dx big)^{1/p}.$$

Show that for $pneq 2$, this norm is not induced by a scalar product.

My method of trying to prove this was to prove a contradiction to the parallelogram rule,

$$ |f+g|^{2}_{p} + |f-g|^{2}_{p} = 2|f|^{2}_{p} + 2|g|^{2}_{p}, tag{$1$}$$

where $f,g in C^{0}([a,b])$.

So I defined the following functions;

$$f(x):=frac{a+b}{2}-x$$

$$g(x) := begin{cases}frac{a+b}{2}-x, for a leq x le frac{a+b}{2}. \
x-frac{a+b}{2}, for frac{a+b}{2} < x le b end{cases}$$

which gives

$$f(x)+g(x) = begin{cases} a+b-2x, & for ale x le frac{a+b}{2}. \ 0, & for frac{a+b}{2} < x le bend{cases}$$

$$f(x)-g(x) = begin{cases} 0, & for a le x le frac{a+b}{2}. \
2x - (a+b), & for frac{a+b}{2} < x le b end{cases}$$

Then I proceeded to calculate each term of the parallelogram rule,

$$|f+g|^{2}_{p} = bigg( int^{frac{a+b}{2}}_{a}|a+b-2x|^{p}bigg)^{2/p} = frac{(b-a)^{frac{2(p+1)}{p}}}{(2(p+1))^{2/p}} $$

$$ |f-g|^{2}_{p} = bigg( int_{frac{a+b}{2}}^{b}|2x- (a+b)|^{p}bigg)^{2/p} = frac{(b-a)^{frac{2(p+1)}{p}}}{(2(p+1))^{2/p}}$$

$$2|f|^{2}_{p} = 2 bigg( int^{b}_{a}| frac{a+b}{2}-x|^{p} dx bigg)^{2/p} = 2 cdot frac{2^{2/p}(frac{b-a}{2})^{frac{2(p+1)}{p}}}{(p+1)^{2/p}} $$

$$begin{align}2 |g|^{2}_{p} & = 2 bigg(int^{frac{a+b}{2}}_{a} |frac{a+b}{2} - x|^{p} dx + int^{b}_{frac{a+b}{2}}|x- frac{a+b}{2}|^{p} dxbigg)^{2/p} \ & =2 cdot frac{2^{2/p}(frac{b-a}{2})^{frac{2(p+1)}{p}}}{(p+1)^{2/p}} end{align}$$

Plugging into $(1)$ we then get
$$2 cdot frac{(b-a)^{frac{2(p+1)}{p}}}{(2(p+1))^{2/p}} = 4 cdot frac{2^{2/p}(frac{b-a}{2})^{frac{2(p+1)}{p}}}{(p+1)^{2/p}}$$
which simplifies quite nicely to

$$2^{p} = 4.$$

So the equality only holds for $p = 2$.

Is what i've done correct? is there another way of proving the question which is better?

Your idea looks fine. However, here's a simpler approach: Let $f = chi_A$ and $g = chi_B$ be the indicator functions of two disjoint sets*. Then

$$|f + g|_p^2 + |f - g|_p^2 = 2 (|A| + |B|)^{2/p}$$

by a direct calculation. On the other hand,

$$2 |f|_p^2 + 2|g|_p^2 = 2 (|A|^{2/p} + |B|^{2/p}).$$

This would imply that for all real numbers $a, b ge 0$ we have

$$(a + b)^{2/p} = a^{2/p} + b^{2/p}.$$

For any $a, b$ which are both positive, this implies $p = 2$ as a consequence of Jensen's inequality. For a specific example, $a = b = 1$ implies $2^{2/p} = 2$, so $p = 2$.

*The point of this answer is that many integral inequalities can be studied purely from the point of view of testing against sets. Although $chi_A$ and $chi_B$ aren't continuous, they can be approximated arbitrarily well in $L^p$ by smooth functions.

Here is a similar idea with simpler computation. Define
$$f(x) = begin{cases} left(frac{a+b}2-xright)^{1/p},&text{ if } x in left[a, frac{a+b}2right] \
0,&text{ if } x in left[frac{a+b}2,bright]
end{cases}$$

$$g(x) = begin{cases} 0,&text{ if } x in left[a, frac{a+b}2right] \
left(x-frac{a+b}2right)^{1/p},&text{ if } x in left[frac{a+b}2,bright]
end{cases}$$

Notice that $f$ and $g$ have disjoint supports so $$f(x)+g(x) = left|frac{a+b}2-xright|^{1/p}$$ and $$f(x)-g(x) = operatorname{sign}left(frac{a+b}2-xright)left|frac{a+b}2-xright|^{1/p}$$

We get
$$|f|_p^2 = left(int_{left[a, frac{a+b}2right]} left(frac{a+b}2-xright),dxright)^{2/p} = left[frac{(b-a)^2}8right]^{2/p}$$
Similarly we also see $|g|_p^2 = left[frac{(b-a)^2}8right]^{2/p}$, $|f+g|_p^2 = left[frac{(b-a)^2}4right]^{2/p}$ and $|f-g|_p^2 = 0$.

Therefore
$$left[frac{(b-a)^2}4right]^{2/p} = |f+g|_p^2 = 4|f_p|^2 = 4left[frac{(b-a)^2}8right]^{2/p}$$

or $2 = 4^{p/2}$. This implies $p=2$.