Is $\langle f,g\rangle$ an inner product on $W$?

[Chris L T521]

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]

This week's question was correctly answered by Sudharaka. You can find his solution below:

Spoiler

\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 \(W\).