# Verify that this is an inner product.

On the space $C^1([0,1])$ I want to show that $(f,g)=\int_0^1(f\overline g+f'\overline {g'})$, where the overline denotes the complex conjugate, defines an inner product. Below is my working for the parts I am most unsure of, and I would like to check if what I have done is correct.

First we establish that $(f,g)$ is well defined. For $t\in[0,1]$ consider,

$$(f,g)=\int_0^1(f\bar g+f'\overline {g'})(t)dt$$

$$=\int_0^1f(t)\overline{g(t)}dt+\int_0^1f'(t)\overline{g'(t)}dt$$

Since $f,g\in C^1([0,1])$, and by the fact that the product of continuous functions is itself continuous, then both integrations are meaningful, and thus, $(f,g)\in\mathbb R$.

Secondly, we establish conjugate symmetry. This is the criterion of an inner product that I am, perhaps, most unsure of. We have that,

$$(f,g)=\int_0^1f(t)\overline{g(t)}dt+\int_0^1f'(t)\overline{g'(t)}dt$$

We use the property that for any $a\in\mathbb R,\,\overline a=a$. For all $t\in[0,1]$, the products $f(t)\overline {g(t)}$ and $f'(t)\overline{g'(t)}$ define real numbers. As such, $\overline {f(t)\overline {g(t)}}=\overline {f(t)}g(t)$ and $\overline{f'(t)\overline{g'(t)}}=\overline{f'(t)}g'(t)$. Thus, we have that,

$$=\int_0^1\overline{f(t)}{g(t)}dt+\int_0^1\overline{f'(t)}g'(t)dt$$

And by the commutativity of multiplication of real numbers,

$$=\int_0^1g(t) \overline{f(t)}dt+\int_0^1g'(t)\overline{f'(t)}dt$$

$$=\overline{(g,f)}$$

Is this correct, or is there anything I could improve upon? I'm fairly confident in proving that $(f,f)\ge0$ and that $(f,f)=0\iff f=0$.

Your reasoning is correct, but it can be simplified. Define, for continuous functions $f$ and $g$ on $[0,1]$ (with complex values), $$(f,g)_0=\int_0^1 f(t)\overline{g(t)}\,dt$$ Then you clearly have, for $f,g\in C^1([0,1])$, $$(f,g)=(f,g)_0+(f',g')_0$$ Both terms on the right hand side are well defined, because by assumption $f'$ and $g'$ are continuous. Now it is easy to prove that $$\bar{f}^{\,\prime}=\overline{f'}$$ (the derivative of the conjugate is the conjugate of the derivative). So you just need to prove that $$(g,f)_0=\overline{(f,g)}$$ for $f$ and $g$ continuous.

What about the last part? First prove that $(f,f)_0=0$ implies $f=0$ for $f$ continuous. Then $$(f,f)=(f,f)_0+(f',f')_0$$ and, as both terms in the right hand side are $\ge0$, the condition $(f,f)=0$ implies $(f,f)_0=0$.

• With regards to proving conjugate symmetry I'm a little hazy on whether what I have is correct, or if it is cutting any corners. Looking what I had again, I reconsidered it; in order that the norm this inner product induces can rightfully be called a norm, we require that $(x,x)=\overline{(x,x)}$. Hence, $\overline{g(t)}f(t)=\overline{\overline{g(t)}f(t)}=g(t)\overline{f(t)}$, and similarly for the other part, using that result on the derivative of the conjugate you mentioned. Is this more formally appropriate than what I provided above, or is it still too "hand wavy"? Feb 5, 2018 at 6:36
• @JeremyJeffreyJames You’re indeed cutting a corner, because you’re not mentioning $\bar f^{\,\prime}=\overline{f’}$, bur that’s all. Oh, and $(f,g)\in\mathbb{C}$. Feb 5, 2018 at 8:02

Perhaps you should say both integrals $\displaystyle\int_{0}^{1}f(t)\overline{g(t)}dt,\displaystyle\int_{0}^{1}f'(t)\overline{g'(t)}dt$ exist, after that the integral defined in $(f,g)$ exists.

• They aren't real in general... Feb 4, 2018 at 22:18
• Existence is the main point, sorry. Feb 4, 2018 at 22:20
• I'm sorry if this is obvious, but am I not only able to say that these integrals are real after splitting it as I done between lines one and two above? Feb 4, 2018 at 22:23
• No, no, they are not real numbers, but you have to say they exist first. Feb 4, 2018 at 22:24
• Strictly speaking you cannot do that, for $\displaystyle\int_{-\infty}^{\infty}xdx$ and $\displaystyle\int_{-\infty}^{\infty}-xdx$ do not exist but the sum $\displaystyle\int_{-\infty}^{\infty}0dx=0$ exists. Feb 4, 2018 at 22:25