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$$\int_{-1}^{1} \overline{f(x)}g(x) \frac{1}{1+x^8} dx$$

Verify that this integral defines a scalar product in $L^2(-1,1)$ between $f(x)$ and $g(x)$. Give then an example of a $f(x)$ for which $\int_{-1}^{1} |f(x)|^2 \frac{1}{1+x^8} dx < \infty $ that it's not in $L^2(-1,1)$.

For the first demonstration, I think it's enough to state that:

1. $\langle f,g\rangle$ = $\overline{\langle g,f\rangle}$

$\overline{\langle g,f \rangle}= \overline{\int \frac{1}{1+x^8} \overline{g(x)}f(x)dx} = \int \frac{1}{1+x^8} \overline{f(x)}g(x)dx$.

2. $\langle f, \alpha g+\beta h\rangle = \int_{-1}^{1}\frac{1}{1+x^8} \overline{f(x)} (\alpha g(x) + \beta h(x)) dx = \alpha\langle f,g \rangle+\beta\langle f,h\rangle$.

3. $\langle f,f\rangle = \int_{-1}^{1} \frac{1}{1+x^8} |f(x)|^2dx \geq 0$.

Should I put something else?

I could not think of a function for the last request.

mechanodroid
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ennedes
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1 Answers1

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Such a function $f$ does not seem to exist.

Notice that

$$x \in [-1,1] \implies x^8 \in [0,1] \implies 1+x^8 \in [1,2] \implies \frac{1}{1+x^8} \in \left[\frac12, 1\right]$$

Which means:

$$\frac12\int_{-1}^{1}\left|f(x)\right|^2\frac{1}{1+x^8}\,dx \le \int_{-1}^{1}\left|f(x)\right|^2\,dx \leq \int_{-1}^{1}\left|f(x)\right|^2\frac{1}{1+x^8}\,dx$$

So $$\int_{-1}^{1}\left|f(x)\right|^2\,dx \text{ is finite } \iff \int_{-1}^{1}\left|f(x)\right|^2\frac{1}{1+x^8}\,dx \text{ is finite }$$

Your verification that $\langle \cdot, \cdot \rangle$ is an inner product is correct, but you have to note that $\langle \cdot, \cdot \rangle$ is well-defined for functions in $L^2(-1,1)$, meaning $\langle f, g \rangle < +\infty$, and verify the property $\langle f,f\rangle = 0 \implies f = 0$. Both easily follow from the above inequality.

mechanodroid
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