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


1 Answer 1


$T(y - \frac{T(y)}{T(x_0)}x_0)=Ty - \frac{T(y)}{T(x_0)}T(x_0)$ 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 $x_0 \neq 0$).

  • $\begingroup$ 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. $\endgroup$
    – CBBAM
    Feb 5 at 0:12
  • $\begingroup$ 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? $\endgroup$ Feb 5 at 0:19
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    $\begingroup$ @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. $\endgroup$ Feb 5 at 2:22
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    $\begingroup$ @CCBAM Where did your equation $\ x_0=\alpha^{-1}x $ come from? $\endgroup$ Feb 5 at 2:49
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    $\begingroup$ @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". $\endgroup$ Feb 5 at 6:33

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