# Problem of the Week #18 - July 30th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem.

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Problem: Let $W$ be the vector space of all differentiable real-valued functions on the interval $[0,1]$. For $f,g\in W$, define

$\langle f,g\rangle=\int_0^1 f(x)g(x)\,dx + \int_0^1 f^{\prime}(x)g^{\prime}(x)\,dx.$

Prove that $\langle f,g\rangle$ is an inner product on $W$.

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#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka. You can find his solution below:

\begin{eqnarray}

\langle f,g\rangle&=&\int_0^1 f(x)g(x)\,dx + \int_0^1 f^{\prime}(x)g^{\prime}(x)\,dx\\

&=&\int_0^1 g(x)f(x)\,dx + \int_0^1 g^{\prime}(x)f^{\prime}(x)\,dx\\

&=&\langle g,f\rangle~~~~~~~~~~~(1)

\end{eqnarray}

\begin{eqnarray}

\langle f+h,g\rangle&=&\int_0^1 (f+h)(x)g(x)\,dx + \int_0^1 (f+h)^{\prime}(x)g^{\prime}(x)\,dx\\

&=&\int_0^1 f(x)g(x)\,dx + \int_0^1 f^{\prime}(x)g^{\prime}(x)\,dx+\int_0^1 h(x)g(x)\,dx + \int_0^1 h^{\prime}(x)g^{\prime}(x)\,dx\\

&=&\langle f,g\rangle+\langle h,g\rangle~~~~~~~~~~~(2)

\end{eqnarray}

\begin{eqnarray}

\langle kf,g\rangle&=&\int_0^1 kf(x)g(x)\,dx + \int_0^1 kf^{\prime}(x)g^{\prime}(x)\,dx\\

&=&k\left[\int_0^1 f(x)g(x)\,dx + \int_0^1 f^{\prime}(x)g^{\prime}(x)\,dx\right]\\

&=&k\langle f,g\rangle~~~~~~~~~~~(3)

\end{eqnarray}

\begin{eqnarray}

\langle f,f\rangle&=&\int_0^1 [f(x)]^2\,dx + \int_0^1 [f^{\prime}(x)]^2\,dx\\

&=&\int_0^1 \left\{[f(x)]^2+[f^{\prime}(x)]^2\right\}\,dx\\

&\geq& 0~~~~~~~~~~~(4)

\end{eqnarray}

$$\mbox{If, }\langle f,f\rangle=0$$

$\int_0^1 \left\{[f(x)]^2+[f^{\prime}(x)]^2\right\}\,dx=0$

$\Rightarrow [f(x)]^2+[f^{\prime}(x)]^2=0$

$\Rightarrow f(x)=0~\forall~x\in[0\,,\,1]$

$$\mbox{Conversely if, }f(x)=0~\forall~x\in[0\,,\,1]$$

$\langle f,f\rangle=\langle 0,0\rangle=0$

$\therefore \langle f,f\rangle=0\Leftrightarrow f(x)=0~\forall~x\in[0\,,\,1]~~~~~~~~~~~~(5)$

Hence, $$\langle f,g\rangle$$ is an inner product on $$W$$.

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