# Why does the kernel and a one dimensional subspace span the Hilbert space when proving the Riesz lemma?

In Reed & Simon's text on functional analysis,, they state the Riesz lemma as

For each $$T \in H^*$$, there is a unique $$y_T \in H$$ such that $$T(x) = (y_T, x)$$ for all $$x \in H$$. In addition $$\|Y_T\|_H = \|T\|_{H^*}$$.

Their (paraphrased) proof is:

Let $$N = \{x \in H : T(x) = 0\}$$ and note that $$N$$ is closed (since it is the preimage of a closed set and $$T$$ is continuous). Assume $$N \neq H$$, so there is a nonzero vector $$x_0 \in N^\perp$$. Let $$Y_T = \overline{T(x_0)}\|x_0\|^{-2}x_0.$$ Then, if $$\alpha x_0 = x \in N$$, $$T(x) = (y_T, \alpha x_0)$$. Since the function $$T(\cdot)$$ and $$(y_T, \cdot)$$ are linear and agree on $$N$$ and $$x_0$$, they must agree on the space spanned by $$N$$ and $$x_0$$.

Now this is the part I do not understand:

But $$N$$ and $$x_0$$ span $$H$$ since every element $$y \in H$$ can be written $$y = \Big(y - \frac{T(y)}{T(x_0)}x_0\Big) + \frac{T(y)}{T(x_0)}x_0.$$ Thus $$T(x) = (y_T, x)$$ for all $$x \in H$$.

They then go on to prove uniqueness and equality of the two norms, but I understand those so I have omitted them.

How does $$y$$ being written in that form imply that $$N$$ and $$x_0$$ span $$H$$? I assume it somehow implies that $$y - \frac{T(y)}{T(x_0)}x_0 \in N$$, but I do not see how. Also, since we assumed $$x \in N$$, and $$x = \alpha x_0$$, by linearity, isn't $$x_0$$ also in $$N$$?

EDIT: $$T(x) = T(\alpha x_0) = \alpha T(x_0) = \overline{(T(x_0)}\|x_0\|^{-2}x_0, \alpha x_0) = (y_T, \alpha x_0)$$

$$T(y - \frac{T(y)}{T(x_0)}x_0)=Ty - \frac{T(y)}{T(x_0)}T(x_0)$$ by linearity of $$T$$ (since $$\frac{T(y)}{T(x_0)}$$ is a scalar).
($$x_o \notin N$$ since $$x_0 \in N^{\perp}$$ and $$x_0 \neq 0$$).
• Thank you, how does the first line imply that $N$ and $x_0$ span $H$? Regarding $x_0$, we assumed that $x \in N$ which a closed subspace, so shouldn't it also contain $x_0$? I know this is incorrect because Reed & Simon assume that $x_0 \in N^\perp$, but I do not see where I have made a mistake. Feb 5 at 0:12
• I do not understand the sentence: Then, if $\alpha x_0 = x \in N$, $T(x) = (y_T, \alpha x_0)....$. $\alpha x_0 = x \in N$ is possible only when $\alpha =0$. Are you sure you are quoting the proof correctly? Feb 5 at 0:19
• @CBBAM it shows that $\ T\left(y-\frac{T(y)}{T\big(x_0\big)}x_0\right)=0\$ and hence that $\ y-\frac{T(y)}{T\big(x_0\big)}x_0\in N\$. Therefore, the equation $$y=\left(y-\frac{T(y)}{T\big(x_0\big)}x_0\right)+\frac{T(y)}{T\big(x_0\big)}x_0$$ expresses $\ y\$ as the sum of an element of $\ N\$ and a multiple of $\ x_0\$, and unless $\ T\$ is identically zero (in which case the theorem is trivial) there must exist a non-zero $\ x_0\in N^\perp\$. Why do you think that the closedness of $\ N \$ implies that $\ x_0\in N\$? It doesn't. Feb 5 at 2:22
• @CCBAM Where did your equation $\ x_0=\alpha^{-1}x$ come from? Feb 5 at 2:49
• @CBBAM Your "paraphrase" of Reed and Simon's text is based on a misunderstanding of it. What it actually says is \begin{align} &\text{First, if }\ x\in\mathcal{N},\text{ then } \ T(x)=0=(y_T,x).\text{ Further, if}\\ &x=\alpha x_0,\text{ then }\dots \end{align} The $\ x\$ in the second sentence isn't the same one as in the preceding sentence. The two sentences are making two different hypotheses on the value of $\ x\$. It becomes clearer if you replace "Further" at the beginning of the second sentence with "On the other hand". Feb 5 at 6:33