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My problem is the following:

All functions are assumed to be real-valued. Determine $\dim(\ker A)$ and $\dim(\mathrm{coker}A)$ when $A = \frac{d}{dx} :X \to Y$ and $X = \{u ∈ C^1([0,1]): u(0) = u(1), u′(0) = u′(1)\},~~ Y = \{f ∈ C^0([0,1]): f(0) = f (1)\}$.

Intuitively, I would say that $\dim(\ker)=1$ since the operator we are considering is the differential operator $\frac{d}{dx}$ and polynomials whose derivative is equal to zero are precisely the constant ones.

Concerning $\dim(\mathrm{coker}A)$, I am a bit lost. Does anyone know how to find it?

Sebastiano
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marg
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1 Answers1

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Consider $f \in Y$ and let $u : [0,1] \to \mathbb{R}$ be given by $$u(x) = \int_{0}^x f(t) \, dt.$$

Then $u(0) = 0$ and $u(1) = \int_0^1 f(t) \, dt$. Furthermore, $u'(0) = f(0) = f(1) = u'(1)$. Hence, we see that $u \in X$ if and only if $\int_0^1 f(t) \, dt = 0.$ Hence, $$\text{Im}(A) = \bigg\{ f \in Y \, \bigg| \, \int_0^1 f(t) \, dt = 0 \bigg\}.$$

We claim that the equivalence class of the function which is identically $1$, also denoted $1$, gives a basis for $\text{coker}A$. Consider any element in $\text{coker}A$ and choose a representative $f \in Y$. Let $c = \int_0^1 f(t) \, dt$. Then $[f] = c 1$ since $$\int_0^1 (f(t) - c) \, dt = 0.$$

Hence, $\dim( \text{coker} A) = 1$.

As you have pointed out yourself, $\dim(\ker A) = 1$ since the functions whose derivative is $0$ are precisely the constant ones. Note that this does not just hold for polynomials.

Thusle Gadelankz
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