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Let $E$ be a vector space and $u \in L(E)$. For the following case, determine whether $u$ has an adjoint with respect to the given bilinear form $\phi (x,y)$.

$E = \{ f: \mathbb{R} \to \mathbb{R}: f $ is a polynomial $\}$,

$\phi(f,g) = \int_0^1 f(t)g(t) dt$

$u(f) = \int_0^1 \frac{f(t)}{\sqrt{t}}dt$

To check if there is an adjoint for u, i have to check for the following:

$(u(f), g) = (f, u'(g))$

So:

$(u(f), g) = (\int_0^1 \frac{f(t)}{\sqrt{t}}dt, g) = \int_0^1 (\int_0^1 \frac{f(t)}{\sqrt{t}}dt)g(t) dt$

and

$(f,u'(g)) = \int_0^1f(t)u'(g)dt$.

Now I will have to check if there exists a $u'$ s.t.

$\int_0^1 (\int_0^1 \frac{f(t)}{\sqrt{t}}dt)g(t) dt$ = $\int_0^1f(t)u'(g)dt$.

How can I do that?

karnan
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1 Answers1

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Note that as you define $u(f)$, that is a constant.

Your mistake is that when you compute $\phi(f,g)$, you use the same variable $t$ for two different purposes.

Instead you should have something like

$\int_0^1 u(f)(t)g(t)dt=\int_0^1 [\int_0^1 \frac{f(t')}{\sqrt{t'}} dt']g(t)dt=\int_0^1 \frac{f(t')}{\sqrt{t'}} dt' \cdot\int_0^1g(t)dt$

Oriol B
  • 181
  • Is $u(f)$ a constant because it depends on $f$? Then you choose a fixed variable $t'$? And in the last step, is that e general rule that allows us to split it in this way? – karnan Apr 17 '20 at 14:08
  • Note that your inner product $\phi$ is defined over $t$. When computing the inner product between $u(f)$ and $g$, it turns out that $u(f)$ is a scalar and $g$ a function of $t$. The last step follows from the fact that there is no $t$ in the first integral nor $t'$ in the second. If it's easier for you just let $u(f):=a\in \mathbb{R}$. Then it's clear that $\int a \cdot g(t) dt = a \cdot \int g(t) dt$ isn't it? – Oriol B Apr 17 '20 at 17:32
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    This made it clear - thanks! But how can I compute $(f,u'(g))$ without knowing what $u'$ actually is? – karnan Apr 18 '20 at 08:19
  • In this case there is no adjoint, because $u(f)$ is not a member of the space of functions on [0,1]. I strongly recommend this book. In particular you can take a look at example 2.19 and theorem 2.21. With the notation of this latter, note that in this case $u: H_0=\mathcal{L}([0,1]) \to H_1=\mathbb{R}$, and $g \in H_0$ but $g\notin H_1$ as you need. – Oriol B Apr 18 '20 at 10:28