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,

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


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,


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,


And by the commutativity of multiplication of real numbers,

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


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


2 Answers 2


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 $(f,f)_0=0$.

  • $\begingroup$ 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"? $\endgroup$ Feb 5, 2018 at 6:36
  • $\begingroup$ @JeremyJeffreyJames You’re indeed cutting a corner, because you’re not mentioning $\bar f^{\,\prime}=\overline{f’}$, bur that’s all. Oh, and $(f,g)\in\mathbb{C}$. $\endgroup$
    – egreg
    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.

  • $\begingroup$ They aren't real in general... $\endgroup$
    – user251257
    Feb 4, 2018 at 22:18
  • $\begingroup$ Existence is the main point, sorry. $\endgroup$
    – user284331
    Feb 4, 2018 at 22:20
  • $\begingroup$ 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? $\endgroup$ Feb 4, 2018 at 22:23
  • $\begingroup$ No, no, they are not real numbers, but you have to say they exist first. $\endgroup$
    – user284331
    Feb 4, 2018 at 22:24
  • $\begingroup$ 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. $\endgroup$
    – user284331
    Feb 4, 2018 at 22:25

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