Show that the $L^{p}$ norm $\|f\|_{L^{p}} := \big( \int^{b}_{a} |f(x)|^p\big)^{1/p}$ is not induced by a scalar product for $p \neq 2$.

Question

On $X = C^0\big([a,b]\big)$, 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 $p\neq 2$, 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

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

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

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

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

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

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

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

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?

Answer 1

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 $p = 2$.

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*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.

Answer 2

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

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

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

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

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

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